5 Applications of Definite Integrals

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1 5 Applictions of Definite Integrls The previous chpter introduced the concepts of definite integrl s n re nd s limit of Riemnn sums, demonstrted some of the properties of integrls, introduced some methods to compute vlues of definite integrls, nd begn to exmine few of their uses. This chpter focuses on severl common pplictions of definite integrls. An obvious gol of the chpter is to enble you to use integrtion when you encounter these prticulr pplictions lter in mthemtics or in other fields. A deeper gol is to illustrte the process of going from problem to n integrl, process much broder thn these prticulr pplictions. If you understnd the process, then you cn understnd the use of integrls in mny other fields nd cn even develop the integrls needed to solve problems in new res. Another gol is to give you dditionl prctice evluting definite integrls. Ech section in this chpter follows the sme bsic formt. First we describe problem nd present some bckground informtion. Then we pproximte the solution to the bsic problem using Riemnn sum. An exct nswer comes from tking limit of the Riemnn sum, nd we get definite integrl. After looking t severl exmples of the sme bsic ppliction, we will exmine some vritions. 5. Volumes by Slicing The previous chpter emphsized geometric interprettion of definite integrls s res in two dimensions. This section emphsizes nother geometricl use of integrtion, computing volumes of solid three-dimensionl objects such s those shown in the mrgin. Our bsic pproch will involve cutting the whole solid into thin slices whose volumes we cn pproximte, dding the volumes of these slices together (to get Riemnn sum), nd finlly obtining n exct nswer by tking limit of those sums to get definite integrl.

2 39 pplictions of definite integrls The Building Blocks: Right Solids A right solid is three-dimensionl shpe swept out by moving plnr region A some distnce h long line perpendiculr to the plne of A (see mrgin). We cll the region A fce of the solid nd use the word right to indicte tht the movement occurs long line perpendiculr t right ngle to the plne of A. Two prllel cuts produce one slice with two fces: A slice hs volume, nd fce hs re. Exmple. Suppose fine, uniform mist is suspended in the ir nd tht every cubic foot of mist contins. ounces of wter droplets. If you run 5 feet in stright line through this mist, how wet do you get? Assume tht the front (or cross section) of your body hs n re of 8 squre feet. Solution. As you run, the front of your body sweeps out tunnel through the mist: The volume of the tunnel is the re of the front of your body multiplied by the length of the tunnel: ( volume = 8 ft ) (5 ft) = 4 ft 3 Becuse ech cubic foot of mist held. ounces of wter (which is ( now on you), you swept out totl of 4 ft 3) (. oz ) ft 3 = 8 ounces of wter. If the wter were truly suspended nd not flling, would it mtter how fst you rn?

3 5. volumes by slicing 39 If A is rectngle, then the right solid formed by moving A long line (see mrgin) is 3-dimensionl solid box B. The volume of B is: (re of A) (distnce long the line) = (bse) (height) (width) If A is circle with rdius r meters (see mrgin), then the right solid formed by moving A long line distnce of h meters is right circulr cylinder with volume equl to: (re of A) (distnce long the line) = π (r ft) ] h ft] = πr h ft 3 If we cut right solid perpendiculr to its xis (like slicing block of cheese), then ech fce (cross-section) hs the sme two-dimensionl shpe nd re. In generl, if 3-dimensionl right solid B is formed by moving -dimensionl shpe A long line perpendiculr to A, then the volume of B is defined to be: (re of A) (distnce moved long the line perpendiculr to A) Exmple. Clculte the volumes of the right solids in the mrgin. Solution. The cylinder is formed by moving the circulr bse with cross-sectionl re πr = 9π in distnce ( of 4 inches long line perpendiculr to the bse, so the volume is 9π in ) (4 in) = 36π in 3. The volume of the box is (bse re) (distnce bse is moved) = (8 m ) (3 m) = 4 m 3. We cn lso simply multiply length times width times height to get the sme nswer. The lst shpe consists of two esy right solids with volumes V = ( π 3 ) () = 8π cm 3 nd V = (6)()() = cm 3, so the totl volume is (8π + ) cm cm 3. Prctice. Clculte the volumes of the right solids shown below. Volumes of Generl Solids We cn cut generl solid into slices, ech of which is lmost right solid if the cuts re close together. The volume of ech slice will

4 39 pplictions of definite integrls then be pproximtely equl to the volume of right solid, so we cn pproximte the totl volume of the entire solid by dding up the pproximte volumes of the right-solid slices. First we position n x-xis below the solid shpe (see mrgin) nd let A(t) be the re of the fce formed when we cut the solid perpendiculr to the x-xis where x = t. If P = {x =, x, x,..., x n = b} is prtition of, b] nd we cut the solid t ech x k, then ech slice of the solid is lmost right solid nd the volume of ech slice is pproximtely (re of fce of the slice) (thickness of the slice) A (x k ) x k The totl volume V of the solid is pproximtely equl to the sum of the volumes of the slices: V = (volume of ech slice) A (x k ) x k which is Riemnn sum. The limit, s the mesh of the prtitions pproches (tking thinner nd thinner slices), of the Riemnn sum is the definite integrl of A(x): V A (x k ) x k b A(x) dx Volume by Slices Formul If then S is solid nd A(x) is the re of the fce formed by cut t x mde perpendiculr to the x-xis the volume V of the prt of S sitting bove, b] is: V = b A(x) dx If S is solid (see mrgin), nd A(y) is the re of fce formed by cut t y perpendiculr to the y-xis, then the volume of slice with thickness y k is pproximtely A (y k ) y k. The volume of the prt of S between cuts t y = c nd y = d on the y-xis is therefore: V = d c A(y) dy Whether you slice region with cuts perpendiculr to the x-xis or cuts perpendiculr to the y-xis depends on which slicing method results in slices with cross-sectionl res tht re esiest to compute. Furthermore, slicing one wy my result in definite integrl tht is difficult to compute, while slicing the other wy my result in much esier definite integrl (lthough you often cn t tell which method will result in n esier integrtion process until you ctully set up the integrls).

5 5. volumes by slicing 393 Exmple 3. For the solid shown in the mrgin, the cross-section formed by cut t x is rectngle with bse of inches. () Find formul for the pproximte volume of the slice between x k nd x k. (b) Compute the volume of the solid for x between nd π. Solution. () The volume of slice is pproximtely: (re of the fce) (thickness) = (bse) (height) (thickness) = ( in) (cos(x k ) in) ( x k in) = cos(x k ) x k in 3 (b) If we cut the solid into n slices of equl thickness x nd dd up the pproximte volumes of the slices, we get Riemnn sum n cos(x k ) x k= π cos(x) dx = sin(x) π = so the volume of the solid is in 3. Prctice. For the solid shown in the mrgin, the fce formed by cut t x is tringle with bse of 4 inches. () Find formul for the pproximte volume of the slice between x k nd x k. (b) Use definite integrl to compute the volume of the solid for x between nd. Exmple 4. For the solid shown in mrgin, ech fce formed by cut t x is squre. Compute the volume of the solid. Solution. The volume of slice is pproximtely: (re of the fce) (thickness) = (bse) (thickness) = ( x k ) x k = x k x k Adding up the pproximte volumes of n slices, we get Riemnn sum tht pproximtes the volume of the entire solid: n x k x k k= 4 x dx = 4 x = 8 so the volume of the solid is 8. Exmple 5. Find the volume of the squre-bsed pyrmid shown in the mrgin. Solution. Ech cut perpendiculr to the y-xis yields squre fce, but in order to find the re of ech squre we need formul for the

6 394 pplictions of definite integrls length of one side s of the squre s function of y, the loction of the cut. Using similr tringles (see mrgin), we know tht: s y = 6 s = 6 ( y) = y The rest of the solution is strightforwrd: ] 3 A(y) = (side) = ( y) = 9 ( y + y ) 5 5 so the volume of the solid is: V = = 9 5 = A(y) dy = 5 y y + 3 ] y3 ( + 3 ( y + y ) dy ) ] ( + ) = You my recll from geometry tht the formul for the volume of pyrmid is Bh where B is the re of the bse, which yields the sme 3 result s the definite integrl: ( 6 ) () =. 3 Exmple 6. Form solid with bse tht is the region between the grphs of f (x) = x + nd g(x) = x for x by building squres with heights (sides) equl to the verticl distnce between the grphs of f nd g (see mrgin). Find the volume of this solid. Solution. The re of squre fce is A(x) = (side) nd the length of side is either f (x) g(x) or g(x) f (x), depending on whether f (x) g(x) or g(x) f (x). We cn express this side length s f (x) g(x) but the side length is squred in the re formul, so A(x) = f (x) g(x) = ( f (x) g(x)). Then: V = = = b A(x) dx = ( f (x) g(x)) dx = + x x x 3 + x 4] dx x + x 3 x3 x4 + 5 x5 ] (x + ) x ] dx which results in volume of 6 5. Wrp-Up At first, ll of these volumes my seem overwhelming there re so mny possible solids nd formuls nd different cses. If you concentrte on the differences, things cn indeed seem very complicted.

7 5. volumes by slicing 395 Insted, focus on the pttern of cutting, finding res of fces, volumes of slices, nd dding those volumes. With tht pttern firmly in mind, you cn reson your wy to definite integrl. Try to mke cuts so the resulting fces hve regulr shpes (rectngles, tringles, circles) whose res you cn clculte esily. Try not to let the complexity of the whole solid confuse you. Sketch the shpe of one fce nd lbel its dimensions. If you cn find the re of one fce in the middle of the solid, you cn usully find the pttern for ll of the fces nd then you cn esily set up the integrl. 5. Problems In Problems 5, compute the volume of the solid using the vlues provided in the tble. 3. disk rdius width box bse height width disk dimeter width slice fce re width Five rock slices re embedded with minerl deposits. Use the informtion in the tble to estimte the totl rock volume. box bse height width slice fce re min. re width

8 396 pplictions of definite integrls In Problems 7, represent the volume of ech solid s definite integrl, then evlute the integrl.. For x 3, ech fce is circle with height (dimeter) 4 x m. 7. For x 3, ech fce is squre with height 5 x inches.. For x 4, ech fce is circle with height (dimeter) 4 x m. 8. For x 3, ech fce is rectngle with bse x inches nd height x inches.. For x, ech fce is squre with side extending from y = to y = x For x 4, ech fce is tringle with bse x + m nd height x m. 3. Suppose A nd B re solids (see below) so tht every horizontl cut produces fces of A nd B tht hve equl res. Wht cn we conclude bout the volumes of A nd B? Justify your nswer.

9 5. volumes by slicing 397 In 4 8, represent the volume of ech solid s definite integrl, then evlute the integrl In 9 8, represent the volume of ech solid s definite integrl, then evlute the integrl The bse of solid is the region between one rch of the curve y = sin(x) nd the x-xis, nd crosssections ( slices ) of the solid perpendiculr to the bse (nd to the x-xis) re squres.. The bse of solid is the region in the first qudrnt bounded by the x-xis, the y-xis nd the curve y = cos(x), nd cross-sections ( slices ) of the solid perpendiculr to the bse (nd to the x-xis) re squres.. The bse of solid is the region in the first qudrnt bounded by the x-xis, the y-xis nd the curve y = cos(x), nd slices perpendiculr to the bse (nd to the x-xis) re semicircles.. The bse of solid is the region between one rch of the curve y = sin(x) nd the x-xis, nd slices perpendiculr to the bse (nd to the x-xis) re equilterl tringles. 3. The bse of solid is the region bounded by the prbols y = x nd y = 3 + x x, nd slices perpendiculr to the bse (nd to the x-xis) re: () squres. (b) semicircles. (c) rectngles twice s tll s they re wide. (d) isosceles right tringles with hypotenuse in the bse of the solid.

10 398 pplictions of definite integrls 4. The bse of solid is the first-qudrnt region bounded by the y-xis, the curve y = sin(x) nd the curve y = cos(x), nd slices perpendiculr to the bse (nd to the x-xis) re: () squres. (b) semicircles. (c) rectngles twice s tll s they re wide. (d) isosceles right tringles with hypotenuse in the bse of the solid. 5. The bse of solid is the region bounded by the x-xis, the y-xis nd the prbol y = 8 x, nd slices perpendiculr to the bse (nd to the y-xis) re squres. 6. The bse of solid is the region bounded by the x-xis, the line y = 3 nd the prbol y = 8 x, nd slices perpendiculr to the bse (nd to the y-xis) re squres. 7. The bse of solid is the region bounded below by the line y =, on the left by the line x = nd bove by the prbol y = 8 x, nd slices perpendiculr to the bse (nd to the y-xis) re semicircles. 8. The bse of solid is the region bounded below by the line y =, on the left by the line x = nd bove by the prbol y = 8 x, nd slices perpendiculr to the bse (nd to the x-xis) re semicircles. 9. Clculte () the volume of the right solid in the top figure (b) the volume of the right cone in the bottom figure nd (c) the rtio of the right cone volume to the right solid volume. 3. Clculte () the volume of the right solid in the top figure (b) the volume of the right cone in the bottom figure nd (c) the rtio of the right cone volume to the right solid volume. 3. Clculte () the volume of the right solid in the top figure if ech blob hs re B (b) the volume of the right cone in the bottom figure, using similr blobs to conclude tht the cross-section x units from the y-xis hs re A(x) = B L x nd (c) the rtio of the right cone volume to the right solid volume. 3. Personl clculus : Describe prcticl wy to determine the volume of your hnd nd rm up to the elbow.

11 5. volumes by slicing Prctice Answers ( ). tringulr bse: V = (bse re) (height) = 3 4 (6) = 36 ( ) semicirculr bse: V = (bse re) (height) = π 3 (7) blob -shped bse: V = (bse re) (height) = (8) (5) = 4 in 3. () The bse of ech tringulr slice is 4 nd the height is pproximtely x k so A (x k ) (4) (x k ) = x k nd the volume of the k-th slice is this pproximtely x k x k. (b) Adding up the pproximte volumes of ll n slices yields which is Riemnn sum with limit: x dx = 3 x3 = = 4 3 n= x k x k,

12 4 pplictions of definite integrls 5. Volumes: Disks nd Wshers In the previous section, we computed volumes of solids for which we could determine the re of cross-section or slice. In this section, we restrict our ttention to specil cse in which the solid is generted by rotting region in the xy-plne bout horizontl or verticl line. We cll solid formed in this wy solid of revolution nd we cll the line n xis of rottion. If the xis of rottion coincides with boundry of the region (s in the mrgin figure) then the cross-sections of the region perpendiculr to the xis of rottion will be disks, mking it reltively esy to find formul for the re of cross-section: A(x) = re of disk = π(rdius) The rdius is often function of x, the loction of the cross-section. Exmple. Find the volume of the solid (shown in the mrgin) formed by rotting the region in the first qudrnt bounded by the curve x y = nd the line x = 4 bout the x-xis. Solution. Any slice perpendiculr to the x-xis (nd to the xy-plne) will yield circulr cross-section with rdius equl to the distnce x between the curve y = nd the x-xis, so the volume of the region is given by: V = 4 ] x π dx = 4 π x 4 dx = π 8 x 4 = π or bout 6.8 cubic inches. Sometimes the boundry curve intersects the xis of rottion. Exmple. The region between the grph of f (x) = x nd the horizontl line y = for x is revolved bout the horizontl line y = to form solid (see mrgin). Compute the volume of the solid. Solution. The mrgin figure shows cross-sections for severl vlues of x, ll of them disks. If x, then the rdius of the disk is r(x) = x ; if x, then r(x) = x. We could split up the volume computtion into two seprte integrls, using A(x) = π r(x)] = π x ] for x nd A(x) = π r(x)] = π x ] for x, but: ] ] π x = π ( x ) = π x ]

13 5. volumes: disks nd wshers 4 for ll x so we cn insted compute the volume with single integrl: V = = π π x 5 x5 3 x3 + x ] dx = π ] = π ] x 4 x + dx ] 3 + = 46π 5 or bout Prctice. Find the volume of the solid formed by revolving the region between f (x) = 3 x nd the horizontl line y = bout the line y = for x 3 (see mrgin). Volumes of Revolved Regions ( Disk Method ) If then the region constrined by the grph of y = f (x), the horizontl line y = L nd the intervl, b] is revolved bout the horizontl line y = L then the volume of the resulting solid is: V = b A(x) dx = b π (rdius) dx = b π f (x) L] dx We often refer to this technique s the disk method becuse revolving thin rectngulr slice of the region (tht we might use in Riemnn sum to pproximte the re of the region) results in disk. If the region between the grph of f nd the x-xis (L = ) is revolved bout the x-xis, then the previous formul reduces to: V = b π f (x)] dx Exmple 3. Find the volume generted when the region between one rch of the sine curve (for x π) nd () the x-xis is revolved bout the x-xis nd (b) the line y = is revolved bout the line y =. Solution. () The rdius of ech circulr slice (see mrgin) is just the height of the function y = sin(x): V = = π π π π sin(x)] dx = π x 4 sin(x) ] π = π π sin (x) dx = π ] cos(x) dx π ] π ] = π 4.93

14 4 pplictions of definite integrls (b) Here we use the more generl disk formul with L = : π V = π sin(x) ] π dx = π sin (x) sin(x) + ] dx 4 π = π cos(x) sin(x) + ] dx 4 3 = π 4 x ] π sin(x) + cos(x) 4 ] 3π = π 4 π + ] = 3π 4 π or pproximtely.. Prctice. Find the volume generted when () the region between the prbol y = x (for x ) nd the x-xis is revolved bout the x-xis nd (b) the region between the prbol y = x (for x ) nd the line y = is revolved bout the line y =. 5 Exmple 4. Given tht 5 f (x) dx = 4 nd f (x)] dx = 7, represent the volume of ech solid shown in the mrgin s definite integrl, nd evlute those integrls. Solution. () Here the xis of rottion is y = so: V = 5 π (rdius) dx = 5 (b) Here the xis of rottion is y = so: V = 5 5 π (rdius) dx = 5 π f (x)] dx = π f (x)] dx = 7π 5 = π f (x) + ] dx = π 5 = π π f (x) ( )] dx 5 5 ( f (x)) dx + f (x) dx + = π (5 )] = 9π ] ( f (x)) + f (x) + dx 5 ] dx (c) This is not solid of revolution, even though the cross-sections re disks. Ech disk hs dimeter equl to the function height, so the rdius of ech disk is hlf tht height, nd the volume is: V = 5 The lst one is left for you. ] f (x) π dx = π 5 f (x)] dx = π = 7π 4 Prctice 3. Set up nd evlute n integrl to compute the volume of the lst solid shown in the mrgin.

15 5. volumes: disks nd wshers 43 Solids with Holes Some solids hve holes : for exmple, we might drill cylindricl hole through sphericl solid (such s bll bering) to crete prt for n engine. One pproch involves using n integrl (or using geometry) to compute the volume of the outer solid, then use nother integrl (or geometry) to compute the volume of the hole cut out of the originl solid, nd finlly subtrcting the second result from the first. You should be ble to use this pproch in the next problem. Prctice 4. Compute the volume of the solid shown in the mrgin. A specil cse of solid with hole results from rotting region bounded by two curves round n xis tht does not intersect the region. Exmple 5. Compute the volume of the solid shown in the mrgin. Solution. The fce for slice mde t x hs re: A(x) = re of BIG circle] re of smll circle] = π BIG rdius] π smll rdius] Here the BIG rdius is the distnce from the line y = x + to the x-xis, or R(x) = (x + ) = x + ; similrly, the smll rdius is the distnce from the curve y = x to the x-xis, or r(x) = x =, hence the x cross-sectionl re is: ] A(x) = π x + ] π = π x + x + x ] x The curves intersect where: + 5 x + = x x + x = x + x = x = ± 4( ) = ± 5 Clerly we need x > for this region, so the left endpoint of integrtion must be x = + 5 while the right endpoint is x =, so the volume of the solid is: V = π x + x + x ] ] dx = π 3 = π which simplifies to π 6 3 x3 + x + x + x + 5 ] ( π ) 3 ( ) ] 5.33.

16 44 pplictions of definite integrls The previous Exmple extends the disk method to more generl technique often clled the wsher method becuse big disk with smller disk cut out of the middle resembles wsher ( smll flt ring used with nuts nd bolts). Volumes of Revolved Regions ( Wsher Method ) If then the region constrined by the grphs of y = f (x) nd y = g(x) nd the intervl, b] is revolved bout horizontl line the volume of the resulting solid is: V = b π (R(x)) π (r(x)) ] dx where R(x) represents the distnce from the xis of rottion to the frthest curve from tht xis, nd r(x) represents the distnce from the xis to the closest curve. If r(x) =, the wsher method becomes the disk method. When pplying the wsher method, you should: grph the region drw representtive rectngulr slice of tht region check tht revolving the slice bout the xis of rottion results in wsher locte the limits of integrtion set up n integrl evlute the integrl If you re unble to find n ntiderivtive for the integrnd of your integrl, you cn consult n integrl tble or use numericl methods to pproximte the volume of the solid. You might lso need to use numericl methods to locte where the boundry curves of the region intersect. Exmple 6. Find the volume of the solid generted by rotting the region between the curves y = x nd y = x bout the () x-xis (b) y-xis (c) the line x = (d) the line y = 5. Solution. () The curves intersect where x = x x x = x(x ) =, so the limits of integrtion should involve x = nd x =. Revolving verticl slice of the region with width x bout the x-xis yields wsher with big rdius R(x) = x = x (the line y = x is frthest from the x-xis) nd smll rdius r(x) =

17 5. volumes: disks nd wshers 45 x = x (the prbol is closest to the x-xis when x ). So the volume of the solid is: V = π(x) π(x ) ] dx = π 4x x 4] dx = π 4 3 x3 5 x5 ] 3 = π ] π ] = 64π 5 (b) A verticl slice revolved round the y-xis does not result in wsher so insted we try slicing horizontlly. A horizontl slice of thickness y revolved round the y-xis does result in wsher. The big rdius is the x-distnce from the prbol (where x = y) to the y-xis (where x = ) so R(y) = y. Similrly, the smll rdius is the distnce from the line (where x = y ) to the y-xis (where x = ), so r(y) = y. Becuse the vrible of integrtion is now y, we need y-vlues for the limits of integrtion. At the lower intersection point of the two curves, x = y = ; t the upper intersection point, x = y = x = = 4. So the volume of the solid is: y=4 V = π ( ( y) y ) ] 4 π dy = π y 4 ] y dy = π y= y y3 ] 4 = π 8 6 ] = 8π 3 3 (c) This solid resembles the one from prt (b), except now the rdii re both bigger becuse the region (nd the curves tht form the boundry of the region) re frther wy from the xis of rottion: R(x) = y ( ) = y + nd r(x) = y ( ) = y + : y=4 V = π ( ( y + ) y ) ] π + dy = π = π y= 4 4 (y + ( y + ) ] y 4 y dy = π )] 4 y + y + dy ] 4 3 y 3 4 y3 3 = π 3 6 ] = 6π 3 3 (d) For this solid, slicing the region verticlly s in prt () results in wshers, but here the ner nd fr roles of the curves re reversed: the prbol is frthest wy from y = 5 while the line is closest. The rdii re R(x) = 5 x nd r(x) = 5 x: x= V = π (5 x ) (5 x) ] dx = 36π 5 x= The detils of evluting this definite integrl re left to you. Prctice 5. Find the volume of the solid generted by rotting the region between the curves y = x nd y = x bout the () the line x = 5 (b) the line y = 5.

18 46 pplictions of definite integrls 5. Problems In Problems, find the volume of the solid generted when the region in the first qudrnt bounded by the given curves is rotted bout the x-xis.. y = x, x = 5. y = sin(x), x = π 3. y = cos(x), x = π 3 4. y = 3 x 5. y = 7 x 6. y = 4 9 x 7. y = 5 x 8. x = y 9 3. Use clculus to determine the volume of sphere of rdius r. (Revolve the region bounded by the x-xis nd the top hlf of the circle x + y = r bout the x-xis.) 33. Compute the volume swept out when the top hlf of the ellipticl region bounded by x 5 + y 3 = is revolved round the x-xis (see figure below). 9. x = y. x + y = 4. 9x + 5y = 5. 3x + 5y = 5 In Problems 3 3, compute the volume of the solid formed when the region between the given curves is rotted bout the specified xis. 3. y = x, y = x 4 bout the x-xis 4. y = x, y = x 4 bout the y-xis 5. y = x, y = x 4 bout the y-xis 6. y = x, y = x 4 bout the x-xis 7. y = x, y = x 3 bout the x-xis 8. y = sec(x), y = cos(x), x = π 3 bout the x-xis 9. y = sec(x), y = cos(x), x = π 3 bout the x-xis. y = x, y = x 4 bout y = 3. y = x, y = x 4 bout y = 4. y = x, y = x 4 bout x = 4 3. y = x, y = x 4 bout x = 3 4. y = x, y = x 4 bout x = 5. y = sin(x), y = x, x = bout y = 3 6. y = sin(x), y = x, x = π bout y = 7. y = x, y = 3 x, bout x = 8. y = x, y = 3 x, bout x = 4 9. y = x, y = 3 x, bout y = 3. y = x, y = 3 x, bout y = 3 3. Use clculus to compute the volume of sphere of rdius. (A sphere is formed when the region bounded by the x-xis nd the top hlf of the circle x + y = is revolved bout the x-xis.) 34. Compute the volume swept out when the top hlf of the ellipticl region bounded by x + y b = is revolved round the x-xis. 35. Compute the volume of the region shown below. 36. Compute the volume of sphere of rdius 5 with hole of rdius 3 drilled through its center.

19 5. volumes: disks nd wshers Compute the volume of the region shown in the mrgin. 38. Determine the volume of the doughnut (clled torus, see lower mrgin figure) generted by rotting disk of rdius r with center R units wy from the x-xis bout the x-xis. 39. () Find the re between f (x) = nd the x-xis for x, x x nd x M. Wht is the limit of the re for x M when M? (b) Find the volume swept out when the region in prt () is revolved bout the x-xis for x, x nd x M. Wht is the limit of the volume for x M when M? 5. Prctice Answers. 3 ] 3 π (3 x) ) dx = π ( x) dx = π. () Slicing the region verticlly nd rotting the slice bout the x-xis results in disks, so the volume of the solid is: 3. π x ] dx = π 3 ] x 4 dx = π 5 x5 = 3π 5 (b) Here the slices extend from y = x to y = so the rdius of ech disk is x nd the volume is: 5 π x ] dx = π 5 x + x ] dx = π x x + ] 3 3 x3 = 3π 4 4x + x 4] dx = π 4x 4 3 x3 + ] 5 x5 = 56π 5 π 3 f (x)] dx = π 9 6 f (x) + ( f (x)) ] dx ] = π 9 dx 6 f (x) dx + f (x)] dx = π ] = 9π 4. The volume we wnt cn be obtined by subtrcting the volume of the box from the volume of the truncted cone generted by the rotted line segment. The volume of the truncted cone is: ] ] π x + ] dx = π x + 4x + 4 dx = π 3 x3 + x + 4x = 56π 3 ] while the volume of the box is () = 4 so the volume of the solid shown in the grph is 56π

20 48 pplictions of definite integrls 5. () Slicing the region verticlly nd rotting the slice bout the line x = 5 results in something other thn wsher, so we insted slice the region horizontlly. The slice extends from x = y (frthest from the xis of rottion) to x = y (closest), so the volume of the solid is: 4 ( π 5 y ) π (5 y) ] dy = 3π 3 (b) Slicing the region verticlly nd rotting the slice bout the line y = 5 results in wshers, so the volume is: ( π 5 x ) ] π (5 x) dx = 36π 5

21 5.3 rclength nd surfce re Arclength nd Surfce Are This section introduces two dditionl geometric pplictions of integrtion: finding the length of curve nd finding the re of surfce generted when you revolve curve bout line. The generl strtegy remins the sme: prtition the problem into smll pieces, pproximte the solution on ech smll piece, dd the smll solutions together to form Riemnn sum nd, finlly, tke the limit of the Riemnn sum to get definite integrl. Arclength: How Long Is Curve? In order to better understnd n niml, biologists need to know how it moves through its environment nd how fr it trvels. We need to know the length of the pth it moves long. If we know the object s loction t successive times, then we cn esily clculte the distnces between those loctions nd dd them together to get totl (pproximte) distnce. Exmple. In order to study the movement of whles, mrine biologists implnt smll trnsmitter on selected whles nd trck the loction of whle vi stellite. Position dt t one-hour time intervls over five-hour period ppers in the mrgin figure. How fr did the whle swim during the first three hours? Solution. In moving from the point (, ) to the point (, ), the whle trveled t lest miles. Similrly, the whle trveled t lest ( ) + (3 ) =.4 miles during the second hour nd t lest (4 ) + ( 3) = miles during the third hour. The scientist concluded tht the whle swm t lest = 7 miles during the three-hour period. Prctice. How fr did the whle swim during the entire five-hour time period? It is unlikely tht the whle swm in stright line from loction to loction, so its ctul swimming distnce ws undoubtedly more thn seven miles during the first three hours. Scientists might get better distnce estimtes by recording the whle s position over shorter, five-minute time intervls. Our strtegy for finding the length of curve will resemble the one the scientist used, nd if the loctions re given by formul, then we cn clculte the successive loctions over very short intervls nd get very good pproximtions of the totl pth length. Exmple. Use the points (, ), (, ) nd (3, 9) to pproximte the length of y = x for x 3.

22 4 pplictions of definite integrls Solution. The lengths of the two line segments (see mrgin) re: ( ) + ( ) = + =.4 nd: (3 ) + (9 ) = = so the length of the curve is pproximtely = Prctice. Get better pproximtion of the length of y = x for x 3 by using the points (, ), (, ), (, 4) nd (3, 9). Is your pproximtion longer or shorter thn the ctul length? For curve C (see mrgin), pick some points (x k, y k ) long C nd connect those points with line segments. Then the sum of the lengths of the line segments will pproximte the length of C. We cn think of this s pinning string to the curve t the selected points, nd then mesuring the length of the string s n pproximtion of the length of the curve. Of course, if we only pick few points (s in the mrgin), then the totl length pproximtion will probbly be rther poor, so eventully we wnt lots of points (x k, y k ) close together ll long C. Lbel these points so tht (x, y ) is one endpoint of C nd (x n, y n ) is the other endpoint, nd so tht the subscripts increse s we move long C. Then the distnce between the successive points (x k, y k ) nd (x k, y k ) is: (x k x k ) + (y k y k ) = ( x k ) + ( y k ) nd the totl length of these line segments is simply the sum of the successive lengths: n ( x k ) + ( y k ) k= This summtion does not hve the form g(c k ) x k so it is not Riemnn sum. It is, however, lgebriclly equivlent to n expression very much like Riemnn sum tht will led us to definite integrl representtion for the length of C. If C is given by y = f (x) for x b, so tht y is function of x, we cn fctor ( x k ) from inside the rdicl nd simplify: length of C = n ( x k ) + ( y k ) = k= ( yk n ( x k ) k= + x k ) = n k= n k= ( xk ) + ( y k) ( x k ) ( ) yk + x x k k to get n expression tht looks more like Riemnn sum. The y k x k inside the rdicl should remind you of two things: the slope of ]

23 5.3 rclength nd surfce re 4 line segment (it is, in fct, the slope of the k-th line segment in our dy pproximtion of the curve C) nd derivtive,. If f (x) is both dx continuous nd differentible, then the Men Vlue Theorem gurntees tht there is some number c k between x k nd x k so tht: f (c k ) = f (x k) f (x k ) x k x k = y k x k Review Section 3. if you need to refresh your memory bout the hypotheses nd conclusions of the Men Vlue Theorem. in which cse we cn write: ( ) yk + x x k = k n k= n + f (c k )] x k k= This lst expression is Riemnn sum, so it converges to definite integrl: n + f (c k )] x k k= b + f (x)] dx This definite integrl provides us with formul for the length of curve C given by y = f (x) for x b. Arclength Formul: y = f (x) version In order to be sure tht the sum converges to the integrl, we need the resulting integrnd to be n integrble function. If we require f (x) to be continuous on, b] then the integrnd will be composition of continuous functions, hence continuous, nd we know tht function tht is continuous on closed intervl is integrble. If C is curve given by y = f (x) for x b nd f (x) exists nd is continuous on, b] then the length L of C is given by: L = b + f (x)] dx Exmple 3. Compute the length of y = x for x 3. Solution. Here f (x) = x f (x) = x so the length of this curve is: 3 + x] dx = 3 + 4x dx Unfortuntely we do not (yet) hve technique to find n ntiderivtive of this integrnd, but we cn use numericl methods (such s Simpson s Rule, or clcultor or computer) to determine tht the vlue of the integrl is pproximtely (compre this with the nswers from Exmple nd Prctice ). You will eventully be ble to find n exct vlue for this definite integrl using techniques developed in Section 8.4. Prctice 3. Compute the length of y = x between (, ) nd (4, 6). Prctice 4. Represent the length of one period of y = sin(x) s definite integrl, then find the length of this curve (using technology to pproximte the vlue of the definite integrl, if necessry).

24 4 pplictions of definite integrls More Arclength Formuls Not ll interesting curves re grphs of functions of the form y = f (x). For curve given by x = g(y) we cn mimic the previous rgument (or simply swp x nd y) to rrive t nother rclength formul: Arclength Formul: x = g(y) version If C is curve given by x = g(y) for c y d nd g (y) exists nd is continuous on c, d] then the length L of C is given by: L = d c + g (y)] dy Prctice 5. Compute the length of x = y between (, ) nd (4, 6). Review Section.5 to refresh your memory bout prmetric equtions. The distnce trveled by the prticle nd the length of the curve will be equl s long s the prticle does not trverse ny prt of the curve more thn once on the intervl α t < β. A curve C cn lso be described using prmetric equtions, where functions x(t) nd y(t) give the coordintes of point on the curve specified by prmeter t. We often think of t s time, so tht (x(t), y(t)) represents the position of prticle in the xy-plne t seconds (or minutes or hours) fter time t =. In Section.5, we discovered tht the speed of such prticle t time t is given by: ( ) dx + dt ( ) dy dt To find the distnce the prticle trvels between times t = α nd t = β, we could then integrte this speed function, which would lso tell us the length of the curve. Arclength Formul (Prmetric Version) If C is curve given by x = x(t) nd y = y(t) for α t β nd x (t) nd y (t) exist nd re continuous on α, β] then the length L of C is given by: L = β α ( ) dx + dt ( ) dy dt dt Prctice 6. Compute the length of the prmetric curve given by the functions x(t) = cos(t) nd y(t) = sin(t) for t π. Prctice 7. Compute the length of the prmetric pth given by the functions x(t) = + 3t nd y(t) = 4t for t 3. Ares of Surfces of Revolution In the previous section, we revolved region in the xy-plne bout horizontl or verticl xis to crete solid, then used n integrl to

25 5.3 rclength nd surfce re 43 compute the volume of tht solid. If we insted rotte curve bout n xis, we get surfce, whose surfce re we cn lso compute using n integrl. Just s the integrl formuls for rclength cme from the simple distnce formul, the integrl formuls for the re of surfce of revolution come from the formul for revolving single line segment. If we rotte line segment of length L prllel to line P (see mrgin) bout the line P, then the resulting surfce ( cylinder) cn be unrolled nd lid flt. This flttened surfce is rectngle with re A = π r L. If we rotte line segment of length L perpendiculr to line P nd not intersecting P (see second mrgin figure) bout the line P, then the resulting surfce is the region between two concentric circles (n nnulus ) nd its re is: A = (re of lrge circle) (re of smll circle) = π (r ) π (r ) = π ( ) r + r = π L (r ) (r ) ] = π (r + r ) (r r ) The expression r +r represents the distnce ( of the ) midpoint of the line segment L from the xis of rottion P nd π r +r is the distnce this midpoint trvels when we revolve the line segment bout the xis. It turns out tht this pttern holds when we revolve ny line segment of length L tht does not intersect line P bout the line P (see mrgin): A = (distnce trveled by segment midpoint) (length of line segment) = π (distnce of segment midpoint from line P) L See Problem 5 for proof. Exmple 4. Compute the re of the surfce generted when ech line segment in the mrgin figure is rotted bout the x-xis nd the y-xis. Solution. Line segment B hs length L = nd its midpoint is t (, ), which is unit from the x-xis nd units from the y-xis. When B is rotted bout the x-xis, the surfce re is therefore: π (distnce of midpoint from x-xis) = π() = 4π nd when B is rotted bout the y-xis, the surfce re is: π (distnce of midpoint from y-xis) = π() = 8π Line segment C hs length 5 nd its midpoint is t (7, 4). When C is rotted bout the x-xis, the resulting surfce re is: π (distnce of midpoint from x-xis) 5 = π(4)5 = 4π When C is rotted bout the y-xis, the distnce of the midpoint from the xis is 7, so the surfce re is π(7)5 = 7π.

26 44 pplictions of definite integrls Prctice 8. Find the re of the surfce generted when the grph in the mrgin is rotted bout ech coordinte xis. When we rotte curve C (tht does not intersect line P, s in the second mrgin figure) bout the line P, we lso get surfce. To pproximte the re of tht surfce, we cn use the sme strtegy we used to pproximte the length of curve: select some points (x k, y k ) long the curve, connect the points with line segments, clculte the surfce re of ech rotted line segment, nd dd together the surfce res of the rotted line segments. The rotted line segment with endpoints (x k, y k ) nd (x k, y k ) hs midpoint: ( xk + x (x k, y k ) = k, y ) k + y k nd length: L = (x k x k ) + (y k y k ) = ( x k ) + ( y k ) If we rotte C bout the x-xis, the distnce from the midpoint of the k-th line segment to the x-xis is y k so the surfce re of the k-th rotted line segment will be: ( ) yk + y π (y k ) L = π k ( x k ) + ( y k ) ( ) ] yk + y = π k yk + x x k k If C is given by y = f (x) for x b, nd f (x) is continuous on, b], we cn ppel to the Men Vlue Theorem to find c k with x k < c k < x k nd f (c k ) = y k x so tht our lst expression becomes: k ( ) f (xk ) + f (x π k ) + f (c k )] x k This is not ctully Riemnn sum, becuse the vlues x k, x k nd c k re not ll the sme. Proving tht this sum converges to definite integrl requires some more dvnced techniques. Adding up these pproximtions, we get: n π k= ( f (xk ) + f (x k ) which converges to definite integrl: b ) + f (c k )] x k π f (x) + f (x)] dx tht gives us formul for the surfce re of the revolved curve. Exmple 5. Compute the re of the surfce generted when the curve y = + x for x 3 is rotted bout the x-xis.

27 5.3 rclength nd surfce re 45 Solution. Here f (x) = + x f (x) = x so, using the integrl formul we just obtined: b 3 ( π f (x) + f (x)] dx = π + x ) + 4x dx We do not (yet) know how to find n ntiderivtive for this integrnd, but numericl pproximtion yields result of You will eventully be ble to find n exct vlue for this definite integrl using techniques developed in Section 8.4. More Sufrce Are Formuls If curve C given by y = f (x) for x b is insted rotted bout the y-xis, then the distnce from the midpoint of the k-th line segment to the xis of rottion is x k. Replcing y k with x k in our work on the previous pge yields the formul: b πx + f (x)] dx for the re of the surfce generted by revolving C bout the y-xis (ssuming, s before, tht f (x) is continuous for x b). Exmple 6. Compute the re of the surfce generted when the curve y = + x for x 3 is rotted bout the y-xis. Solution. Here gin f (x) = + x f (x) = x so, using our newest integrl formul: b 3 πx + f (x)] dx = πx + 4x dx We cn find n ntiderivtive of this integrnd using substitution: u = + 4x du = 8x dx du = x dx 8 The integrl limits become u = + 4() = nd u = + 4(3) = 37, so the surfce re is: u=37 π π 37 u du = u π du = ] u = π 37 ] 37 6 u= or pproximtely 7.3. Wrp-Up Developing formuls for the re of surfce generted by rotting curve x = g(y) for c y d (or by prmetric equtions) present little dditionl difficulty. In future chpters, however, we will develop much more generl yet simpler formuls for rclength nd surfce re. While the integrl formuls developed in this section cn be useful, more importntly their development served to illustrte yet gin how reltively simple pproximtion formuls cn led us vi Riemnn sums to integrl formuls. We will see this process gin nd gin. See Problems 48 5.

28 46 pplictions of definite integrls 5.3 Problems. The loctions (in feet, reltive to n ok tree) t vrious times (in minutes) for squirrel spotted in bck yrd pper in the tble below: time north est At lest how fr did the squirrel trvel during the first 5 minutes?. The squirrel in the previous problem trveled t lest how fr during the first 3 minutes? 3. Use the prtition {,, } to estimte the length of y = x between the points (, ) nd (, 4). 4. Use the prtition {,, 3, 4} to estimte the length of y = ( x between the points (, ) nd 4, ). 4 The grphs of the functions in Problems 5 8 re line segments. Clculte ech length () using the distnce formul between two points nd (b) by setting up nd evluting n pproprite rclength integrl. 5. y = + x for x. 6. y = 5 x for x 4, 7. x = + t, y = t for t x = 4t, y = + t for t Clculte the length of y = 3 x 3 for x 4.. Clculte the length of y = 4x 3 for x 9. Very few functions of the form y = f (x) led to integrnds of the form + f (x)] tht hve elementry ntiderivtives. In 4, + f (x) ] ends up being perfect squre, so you cn evlute the resulting rclength integrl using ntiderivtives.. y = x3 3 + for x 5. 4x. y = x4 4 + for x 9. 8x 3. y = x5 5 + for x 5. x3 4. y = x6 6 + for 4 x 5. 6x4 In Problems 5 3, represent ech length s definite integrl, then evlute the integrl (using technology, if necessry). 5. The length of y = x from (, ) to (, ). 6. The length of y = x 3 from (, ) to (, ). 7. The length of y = x from (, ) to (9, 3). 8. The length of y = ln(x) from (, ) to (e, ). ( ) π 9. The length of y = sin(x) from (, ) to 4, ( ) π ( π ) nd from 4, to,.. The length of the ellipse x(t) = 3 cos(t), y(t) = 4 sin(t) for t π.. The length of the ellipse x(t) = 5 cos(t), y(t) = sin(t) for t π.. A robot progrmmed to be t loction x(t) = t cos(t), y(t) = t sin(t) t time t will trvel how fr between t = nd t = π? 3. How fr will the robot in the previous problem trvel between t = nd t =? 4. As tire of rdius R rolls, pebble stuck in the tred will trvel cycloid pth, given by x(t) = R (t sin(t)), y(t) = R ( cos(t)). As t increses from to π, the tire mkes one complete revolution nd trvels forwrd πr units. How fr does the pebble trvel? 5. Referring to the previous problem, s tire with -foot rdius rolls forwrd mile, how fr does pebble stuck in the tire tred trvel? 6. Grph y = x n for n =, 3, nd. As the vlue of n becomes lrge, wht hppens to the grph of y = x n? Estimte the vlue of: lim n x= x= + n x n ] dx

29 5.3 rclength nd surfce re Find the point on the curve f (x) = x for x 4 tht will divide the curve into two eqully long pieces. Find the points tht will divide the segment into three eqully long pieces. 8. Find the pttern for the functions in Problems 4. If y = xn n + Axp, how must A nd p be relted to n? 9. Use the formuls for A nd p from the previous problem with n = 3 nd find new function y = 3 x 3 ] + Ax p so tht + dy dx is perfect squre. 3. Find the surfce re when ech line segment in the figure below is rotted bout the () x-xis nd (b) y-xis. 3. Find the surfce re when ech line segment in the figure bove is rotted bout the line () y = nd (b) x =. 3. Find the surfce re when ech line segment in the figure below is rotted bout the line () y = nd (b) x =. 34. A line segment of length with midpoint (, 5) mkes n ngle of θ with the horizontl. Wht vlue of θ will result in the lrgest surfce re when the line segment is rotted bout the y-xis? Explin your resoning. 35. A line segment of length with one end t (, 5) mkes n ngle of θ with the horizontl. Wht vlue of θ will result in the lrgest surfce re when the line segment is rotted bout the x-xis? Explin your resoning. In Problems 36 43, when the given curve is rotted bout the given xis, represent the re of the resulting surfce s definite integrl, then evlute tht integrl using technology. 36. y = x 3 for x bout the y-xis 37. y = x 3 for x bout the y-xis 38. y = x for x bout the x-xis 39. y = x for x bout the x-xis 4. y = sin(x) for x π bout the x-xis 4. y = x 3 for x bout the x-xis 4. y = sin(x) for x π bout the y-xis 43. y = x for x bout the y-xis 44. Find the re of the surfce formed when the grph of y = 4 x is rotted bout the x-xis: () for x. (b) for x. (c) for x Show tht if thin hollow sphere is sliced into pieces by eqully spced prllel cuts (see below), then ech piece hs the sme weight. (Hint: Does ech piece hve the sme surfce re?) 33. Find the surfce re when ech line segment in the figure bove is rotted bout the () x-xis nd (b) y-xis.

30 48 pplictions of definite integrls 46. Interpret the result of the previous problem for n ornge sliced by eqully spced prllel cuts. 47. A hemisphericl cke with uniformly thick lyer of frosting is sliced with eqully spced prllel cuts. Does everyone get the sme mount of cke? The sme mount of frosting? 48. Devise formul for the re of the surfce generted by revolving the curve x = g(y) for c y d bout the () x-xis nd (b) y-xis. 49. Use the nswer to the previous problem to find the re of the surfce generted by revolving x = e y for y bout () the x-xis nd (b) the y-xis. 5. Devise formul for the re of the surfce generted by revolving the curve given by prmetric equtions x = x(t) nd y = y(t) for α t β bout the () x-xis nd (b) y-xis. 5. Use the nswer to the previous problem to find the re of the surfce generted by revolving the curve given by x = cos(t) nd y = sin(t) for t π bout () the x-xis nd (b) the y-xis. 5. The surfce generted by revolving line segment of length L bout line P (tht does not intersect the line segment) is the frustrum of cone: the surfce tht results from tking lrger cone of rdius r nd removing smller cone of rdius r ( chopping off the top ). We know from geometry tht the surfce re of cone is πrs where r is the rdius of the cone nd s is the slnt height: () If s is the slnt height of the smller cone tht is removed from the bigger cone, show tht: s + L = r L r r (Hint: Use similr tringles.) (b) Show tht the surfce re of the frustrum is: πr (s + L) πr s (c) Show tht this quntity equls: π (r + r ) L (d) Show tht this lst quntity is the product of the distnce trveled by the midpoint nd the length of the line segment. 3-D Arclength If 3-dimensionl curve C (see mrgin) is given prmetriclly by x = x(t), y = y(t) nd z = z(t) for α t β, then we cn esily extend the rclength formul to three dimensions: L = β α ( ) dx + dt ( ) dy + dt ( ) dz dt dt The remining problems in this section use this formul to provide you with preview of clculus in higher dimensions.

31 5.3 rclength nd surfce re Find the length of the helix (see figure) given by x = cos(t), y = sin(t), z = t for t 4π. 54. Find the length of the line segment given by x = t, y = t, z = t for t. 55. Find the length of the curve given by x = t, y = t, z = t 3 for t. 56. Find the length of the stretched helix given by x = cos(t), y = sin(t), z = t for t π. 57. Find the length of the curve given by x = 3 cos(t), y = sin(t), z = sin(7t) for t π. 5.3 Prctice Answers. At lest miles.. L < ctul length x] dx = π + cos(x)] dx x dx Here g(y) = y = y g (y) = y = so the rclength is: y 6 ] 6 + dy = + This dy 5.34 y 4y nswer is the sme s the nswer to Prctice 3. Should tht surprise you? 6. Here x (t) = sin(t) nd y (t) = cos(t) so the rclength is: π π sin(t)] + cos(t)] dt = dt = π The curve in question is circle of rdius. Does the nswer from the integrl formul gree with the nswer you cn obtin using simple geometry? 7. Here x (t) = 3 nd y (t) = 4 so the rclength is: 3 3 3] + 4] dt = 5 dt = The curve is line segment from (4, 4) to (, ). Does the nswer from the integrl formul gree with the nswer you cn obtin using simple geometry? 8. The surfce re of the horizontl segment revolved bout x-xis is π()() = 4π.57 while the surfce re of other segment revolved bout the x-xis is π()( 8) 35.54, so the totl surfce re is pproximtely = 48. squre units. The surfce re of the horizontl segment revolved bout y-xis is π(3)() = π 37.7 while the surfce re of the other segment revolved bout the y-xis is π(5)( 8) 88.86, so the totl surfce re is pproximtely = 6.56 squre units.

32 4 pplictions of definite integrls 5.4 More Work There re so mny possible vritions in work problems tht it is vitl you understnd the process. In Section 4.7 we investigted the problem of clculting the work done in lifting n object using cble. This section continues tht investigtion nd extends the process to hndle situtions in which the pplied force or the distnce or both my vry. The method we used before turns up gin here. The first step is to divide the problem into smll slices so tht the force nd distnce vry only slightly on ech slice. Then we clculte the work done for ech slice, pproximte the totl work by dding together the work for ech slice (to get Riemnn sum) nd, finlly, tke limit of tht Riemnn sum to get definite integrl representing the totl work. Recll tht the work done on n object by constnt force is defined to be the mgnitude of the force pplied to the object multiplied by the distnce over which the force is pplied: work = (force) (distnce) A similr exmple ppered in Section 4.7, but it provides good illustrtion of the process of dividing problem into pieces nd nlyzing ech piece. Exmple. A -pound object is lifted 4 feet from the ground to the top of building using cble tht weighs pound per foot (see mrgin figure). How much work is done? Solution. The work done on the object is simply: W = F d = ( lbs) (4 ft) = 4 ft-lbs For the rope, we cn prtition it (see second mrgin figure) into n smll pieces, ech with length x. Ech smll piece of rope weighs: ( ) lb ( x ft) = x lb ft nd the k-th slice of rope is lifted distnce of (pproximtely) 4 x k feet, so the work done on the k-th slice of rope is (pproximtely): ( ) ( ) W k = F k d k = x lb (4 x k ) ft = (4 x k) x ft-lbs nd the totl work done to lift the rope is therefore: n k= (4 x k) x 4 (4 x) dx Evluting this integrl yields: 4x ] 4 x = 6 8] = 4 ft-lbs so the totl work done to lift the object is = 8 ft-lbs. Prctice. How much work is done lifting 3-pound injured person to the top of 3-foot cliff using stretcher tht weighs pounds nd cble weighing pounds per foot?

33 5.4 more work 4 Work in the Metric System All of the work problems we hve considered so fr mesured force in pounds nd distnce in feet, so tht work ws mesured in footpounds. In the metric system, we often mesure distnce in meters (m) nd force in newtons (N). According to Newton s second lw of motion: force = (mss) (ccelertion) or, more succinctly, F = m. The force in mny work problems is the weight of n object, so the ccelertion in question is the ccelertion due to grvity, denoted by g. Ner se level on Erth, g sec m, lthough the vlue 9.8 is commonly used in computtions. An object with mss of kg would thus hve weight of: ( mg = ( kg) 9.8 m ) sec = 98. kg m sec = 98. N Virtully ll countries other thn the United Sttes long with U.S. scientists nd engineers use the metric system, so you need to know how to solve work problems using metric units. Sir Isc Newton (643 77) not only invented clculus, he formulted the lws of motion nd universl grvittion in physics (mong mny other ccomplishments). Exmple. An object with mss of kg is lifted 4 m from the ground to the top of building using 4-meter cble with mss of kg. How much work is done? Solution. The work done on the object is: ( W = F d = mg d = ( kg) 9.8 m ) sec (4 m) = 394 N-m or 3,94 joules ( joule, bbrevited J, is N-m). The cble hs totl mss kg nd is 4 m long, so it hs liner density of: kg 4 m = kg m We cn prtition the cble into n smll pieces, ech with length x, so ech smll piece of cble hs mss of: ( ) kg ( x m) = x kg m nd thus hs weight of: ( ) ( F = mg = x kg 9.8 m ) sec = 4.95 x kg m sec = 4.95 x N The k-th slice of cble is lifted distnce of pproximtely 4 x k m, so the work done on the k-th slice of cble is: ( ) W k = F k d k = (4.95 x N) (4 x k ) m = 4.95 (4 x k ) x N-m This unit for work is nmed fter nother English physicist, Jmes Prescott Joule (88 889). nd the totl work done lifting the cble is therefore: n 4.95 (4 x k ) x k= (4 x) dx = 394 J so the totl work done to lift the object is = 7848 J. Much of this process should look fmilir. Compre the solution of Exmple to tht of Exmple on the previous pge.

34 4 pplictions of definite integrls Prctice. How much work is done lifting n injured person of mss 5 kg to the top of 3-meter cliff using stretcher of mss 5 kg nd 3-meter cble of mss kg? Lifting Liquids Exmple 3. A col glss (see mrgin figure) hs dimensions given in the mrgin tble. Approximtely how much work do you do when you drink col glss full of wter by sucking it through strw to point 3 inches bove the top edge of the glss? Solution. The tble prtitions the wter into -inch slices : height rdius You might wonder why the displcement is not computed by tking the distnce from the bottom of the strw up to the top of the strw, but when computing work we need to use the net displcement. Wter s density is 6.5 lb oz = ft in 3. The work needed to move ech slice is pproximtely the weight of the slice times the distnce the slice is moved. We cn use the rdius t the bottom of ech slice to pproximte the volume nd then the weight of the slice, nd point hlfwy up ech slice to clculte the distnce the slice is moved. For the top slice: weight = (volume) (density) π (.6 in) ( in) (.5787 oz ) in oz nd the distnce this slice trvels is roughly 3.5 inches, so: For the next slice: W = F d (4.7 oz) (3.5 in) 6.4 oz-in weight = (volume) (density) π (.5 in) ( in) nd the distnce this slice trvels is roughly 4.5 inches, so: W = F d (4. oz) (4.5 in) 8.4 oz-in (.5787 oz ) in 3 4. oz

35 5.4 more work 43 The work for the lst two slices is (.8 oz) (5.5 in) = 9.9 oz-in nd (. oz) (6.5 in) = 4.3 oz-in. The totl work is then sum of the work needed to rise ech slice of wter: (6.4 oz-in) + (8.4 oz-in) + (9.9 oz-in) + (4.3 oz-in) = 59. oz-in or bout.3 ft-lbs. Prctice 3. Approximte the totl work needed to rise the wter in Exmple 3 by using the top rdius of ech slice to pproximte its weight nd the midpoint of ech slice to pproximte the distnce the slice is rised. If we knew the rdius of the col glss t every height, then we could improve our pproximtion by tking thinner nd thinner slices. In fct, we could hve formed Riemnn sum, tken the limit of the Riemnn sum s the thickness of the slices pproched, nd obtined definite integrl. In the next Exmple we do know the rdius of the continer t every height. Exmple 4. Find the work needed to rise the wter in the cone shown below to the top of the strw. In this exmple, both the force nd the distnce vry, nd ech depends on the height of the slice bove the bottom of the cone. Solution. We cn prtition the cone to get n slices of wter. The work done rising the k-th slice is the product of the distnce the slice is rised nd the force needed to move the slice (the weight of the slice). For ny c k in the subintervl y k, y k ], the slice is rised distnce of pproximtely ( c k ) cm. Ech slice is pproximtely right circulr cylinder, so its volume is: If you wnt, you cn choose c k = y k like you did in Prctice 3. π (rdius) y At height y bove the bottom of the cone, the rdius of the cylinder is x = y 3 so t height c k the rdius is 3 c k; the mss of ech slice is To see this, use similr tringles in the right-hnd figure bove: x y = 6 x = y 3

36 44 pplictions of definite integrls In the metric system, grm (bbrevited g ) is defined s the mss of one cubic centimeter of wter, so the density of wter is: g kg =, cm3 m 3 In the g cm sec version of the metric system, the stndrd unit of force is dyne (bbrevited dyn ), which is g-cm sec : N =, dyn In the g cm sec version of the metric system, the stndrd unit of work is clled n erg, which is dyn-cm: J =,, erg therefore: ( (volume) (density) π (rdius) ( y) g ) cm ( ) 3 ( = π 3 c k cm ( y cm) g ) cm 3 = π 9 (c k) y g so the force required to rise the k-th slice is: F k = m k g π 9 (c k) y g] 98 cm ] sec = 9π (c k ) y g-cm sec nd the work required to lift the k-th slice is: W k = F k d k ] 9π (c k ) y dyn ( c k ) cm] = 9π (c k ) ( c k ) y dyn-cm We cn then dd the work done on ll n slices to get Riemnn sum: W n 9π (c k ) ( c k ) y k= y=6 y= 9πy ( y) dy Evluting this integrl is reltively strightforwrd: We integrte from y = to y = 6 becuse the bottom slice of wter is t height of cm nd the top slice of wter is t height of 6 cm. 6 ( W = 9π y y 3) dy = 9π 3 y3 ] 6 4 y4 = 9π 7 34] = 4364π erg or bout 35, 64 erg =.3564 J. Prctice 4. How much work is done drinking just the top 3 cm of the wter in Exmple 4? Exmple 5. The trough shown in the mrgin is filled with liquid weighing 7 pounds per cubic foot. How much work is done pumping the liquid over the wll next to the trough? Solution. As before, we cn prtition the height of the trough to get n slices of liquid (see mrgin figure t top of next pge). To form Riemnn sum for the totl work, we need the weight of typicl slice nd the distnce tht slice is rised. The weight of the k-th slice is: ( (volume) (density) (length) (width) (height) 7 lb ) ft 3 The length of ech slice is 5 feet, nd the height of ech slice is y feet, but the width of ech slice (w k ) vries nd depends on how fr the slice

37 5.4 more work 45 is bove the bottom of the trough (c k ). Using similr tringles on the edge of the trough, we cn observe tht: w k c k = 4 w k = c k so the weight of the k-th slice is therefore: ( ck ) ( (5 ft) ft ( y ft) 7 lb ) ft 3 = 75c k y lb The k-th slice is rised from height of c k feet to height of 6 feet, through distnce of 6 c k feet, so the work done on the k-th slice is: ] ] W k = F k d k 75c k y lb (6 c k ) ft = 75c k (6 c k ) y lb-ft Adding up the work done on ll n slices yields Riemnn sum tht converges to definite integrl: n 75c k (6 c k ) y k= 4 75y(6 y) dy Evluting the integrl is strightforwrd: 4 ( 75 6y y ) dy = 75 3y ] 4 3 y3 = 4 3 or bout 4, 667 ft-lbs. You cn generlly hndle rise the liquid problems by prtitioning the height of the continer nd then focusing on typicl slice. Prctice 5. How much work is done pumping hlf of the liquid over the wll in Exmple 5? We integrte from y = to y = 4 becuse the bottom slice of liquid is t height of feet nd the top slice of liquid is t height of 4 feet. If you cn clculte the weight of typicl slice nd the distnce it is rised, the rest of the steps re strightforwrd: form Riemnn sum, let it converge to definite integrl, nd evlute the integrl to get the totl work. Work Moving n Object Along Stright Pth If you push box long flt surfce (s in the figure below) tht is smooth in some plces nd rough in others, t some plces you only need to push the box lightly nd in other plces you hve to push hrd. If f (x) is the mount of force needed t loction x, nd you wnt to push the box long stright pth from x = to x = b, then we cn prtition the intervl, b] into n pieces,, x ], x, x ],..., x n, b]: The force f (x) discussed here is the minimum force required to counterct the kinetic friction between the box nd the surfce t ny point. You will lern more bout friction in physics nd engineering clsses.

38 46 pplictions of definite integrls The work required to move the box through the k-th subintervl, from x k to x k, is pproximtely: (force) (distnce) f (c k ) (x k x k ) = f (c k ) x k for ny c k in the subintervl x k, x k ]. The totl work is the sum of the work long these n pieces, which is Riemnn sum tht converges to definite integrl: n k= f (c k ) x k b f (x) dx This hs simple geometric interprettion. If f (x) is the force pplied t position x, then the work done to move the object from position x = to position x = b is the re under the grph of f between x = nd x = b (see mrgin). This formul pplies in more generl situtions, s demonstrted in the next Exmple. Exmple 6. If force of 7x pounds is required to keep spring stretched x inches pst its nturl length, how much work will be done stretching the spring from its nturl length (x = ) to five inches beyond its nturl length (x = 5)? Solution. According to the formul we just developed: work = b f (x) dx = 5 7x dx = ] 7 5 x = 75 = 87.5 in-lbs or bout 7.9 ft-lbs. (See mrgin for grphicl interprettion.) Prctice 6. How much work is done to stretch the spring in Exmple 6 from 5 inches pst its nturl length to inches pst its nturl length? The preceding spring exmple is n ppliction of physicl principle discovered by English physicist Robert Hooke (635 73), contemporry of Newton. Hooke s Lw: The force f (x) needed to keep spring stretched (or compressed) x units beyond its nturl length is proportionl to the distnce x: f (x) = kx for some constnt k. In fct, Hooke s Lw holds for most solid objects, t lest for limited forces: Nor is it observble in these bodies only, but in ll other springy bodies whtsoever, whether metl, wood, stones, bked erth, hir, horns, silk, bones, sinews, glss nd the like. Robert Hooke, De Potenti Restitutiv, or Of Spring We cll the k in Hooke s Lw spring constnt. It vries from spring to spring (depending on the mterils nd dimensions of the spring nd even on the temperture of the spring), but remins constnt for ech spring s long s the spring is not overextended or overcompressed. Most bthroom scles use compressed springs nd Hooke s Lw to mesure person s weight.

39 5.4 more work 47 Exmple 7. A spring hs nturl length of 43 cm when hung from ceiling. A mss of 4 grms stretches it to length of 75 cm. How much work is done stretching the spring from length of 63 cm to length of 93 cm? Solution. First we need to use the given informtion to find the vlue of k, the spring constnt. A mss of 4 g produces stretch of = 3 cm. Substituting x = 3 cm nd f (x) = 4 g 98 cm into Hooke s sec Lw f (x) = kx, we hve: 4(98) = k(3) k = The length of 63 cm represents stretch of cm beyond the spring s nturl length, while the length of 93 cm represents 5-cm stretch. The work done is therefore: 5 ] x dx = 8 x = ] = ergs or bout.9 joules. Prctice 7. A spring hs nturl length of 3 inches when hung from ceiling, nd force of pounds stretches it to length of 8 inches. How much work is done stretching the spring from length of 5 inches to length of inches? Lifting Pylod Clculting the work required to lift pylod from the surfce of moon (or ny body with no tmosphere) cn be ccomplished using similr computtion. Newton s Lw of Universl Grvittion sys tht the grvittionl force between two bodies of mss M nd m is: F = GMm x ( ) where G N m is the grvittionl constnt nd x is kg the distnce between (the centers of) the two bodies. If the moon hs rdius of R m nd mss M, the pylod hs mss m nd x mesures the distnce (in meters) from pylod to the center of the moon (so x R), then the totl mount of work done lifting the pylod from the surfce of the moon (n ltitude of, where x = R) to n ltitude of R (where x = R + R = R) is: R R GMm x dx = GMm ] R = GMm x R R + ] = GMm R R

40 48 pplictions of definite integrls Prctice 8. How much work is required to lift the pylod from n ltitude of R m bove the surfce (x = R) to n ltitude of R m? The pproprite res under the force grph (see mrgin) illustrte why the work to lift the pylod from x = R to x = R is much lrger thn the work to lift it from x = R to x = 3R. In fct, the work to lift the pylod from x = R to x = R is.49gmmr, which is less thn the.5gmmr needed to lift it from x = R to x = R. The rel-world problem of lifting pylod turns out to be much more chllenging, becuse the rocket doing the lifting must lso lift itself (more work) nd the mss of the rocket keeps chnging s it burns up fuel. Lifting pylod from moon (or plnet) with n tmosphere is even more difficult: the tmosphere produces friction, nd the frictionl force depends on the density of the tmosphere (which vries with height), the speed of the rocket nd the shpe of the rocket. Life cn get complicted. Power Scottish engineer Jmes Wtt (736 89) devised horsepower to compre the output of stem engines with the power of drft horses. In physics, power is defined s the rte of work done per unit of time. One trditionl mesurement of power, horsepower (bbrevited hp ), originted with Jmes Wtt s determintion in 78 tht horse could turn mill wheel of rdius feet 44 times in n hour while exerting force of 8 pounds. Such horse would trvel: 44 rev hr π() ft rev 6 hr min = 88π 5 ft min nd so it would produce work t rte of: ( ) 88π ft (8 lb) = 368π ft-lb ft-lb min min min which Wtt subsequently rounded to: hp 746 W 33 ft-lb ft-lb = 55 min sec = horsepower The metric unit of power, clled wtt (bbrevited W ) in Wtt s honor, is equivlent to joule per second. Exmple 8. How long will it tke for -horsepower electric pump to pump ll of the liquid in the trough from Exmple 5 over the wll? Solution. Power (P) is the rte t which work (W) is done, so: P = W t or bout 8.5 seconds. t = W 4 P = 3 ft-lbs = hp 4 3 ft-lbs 55 ft-lbs = 8 33 sec sec

41 5.4 more work Problems. A tnk 4 feet long, 3 feet wide nd 7 feet tll (see below) is filled with wter. How much work is required to pump the wter out over the top edge of the tnk? () How much work is done pumping ll of the wter over the top edge of the qurium? (b) How long does it tke for -horsepower pump to empty the tnk? A 4 -horsepower pump? Which pump does more work? (c) If the qurium is only filled to height of 4 feet with se wter, how much work is required to empty it? 5. A cylindricl brrel with rdius of m nd height of 6 m is filled with wter.. A tnk 4 feet long, 3 feet wide nd 6 feet tll is filled with oil with density of 6 pounds per cubic foot. () How much work is needed to pump ll of the oil over the top edge of the tnk? (b) How much work is needed to pump the top 3 feet of oil over the top edge of the tnk? 3. A tnk 5 m long, m wide nd 4 m tll is filled with n oil of density 9 kg/m 3. () How much work is needed to pump ll of the oil over the top edge of the tnk? (b) How much work is needed to pump the top m 3 of oil over the top edge of the tnk? (c) How long does it tke for -wtt pump to empty the tnk? 4. A cylindricl qurium with rdius feet nd height 5 feet (see below) is filled with slt wter (which hs density of 64 pounds per cubic foot). () How much work is done pumping ll of the wter over the top edge of the brrel? (b) How much work is done pumping the top m of wter to point m bove the top edge of the brrel? (c) How long will it tke -horsepower pump to remove hlf of the wter from the brrel? 6. The conicl continer shown below is filled with ots tht weigh 5 pounds per ft 3. () How much work is done lifting ll of the grin over the top edge of the cone? (b) How much work is required to lift the top feet of grin over the top edge of the cone? 7. If you nd friend shre the work eqully in emptying the conicl continer in the previous problem, wht depth of grin should the first person leve for the second person to empty?

42 43 pplictions of definite integrls 8. A trough (see below) is filled with pig slop weighing 8 pounds per ft 3. How much work is done lifting ll the slop over the top of the trough? 9. In the preceding problem, how much work is done lifting the top 4 ft 3 of slop over the top edge of the trough?. The prbolic continer shown below (with height of 4 m) is filled with wter. 3. The sphericl tnk shown bove is filled with wter to depth of m. How much work is done lifting ll of tht wter to the top of the tnk? 4. A student sid, I ve got shortcut for these tnk problems, but it doesn t lwys work. I figure the weight of the liquid nd multiply tht by the distnce I hve to move the middle point in the wter. It worked for the first five problems nd then it didn t. () Does this shortcut relly give the right nswer for the first five problems? (b) How do the continers in the first five problems differ from the others? (c) For which of the continers shown below will the shortcut work? () How much work is done pumping ll of the wter over the top edge of the tnk? (b) How much work is done pumping ll of the wter to point 3 m bove the top of the tnk?. The prbolic continer shown below (with height of m) is filled with wter. 5. All of the continers shown below hve the sme height nd hold the sme volume of wter. () How much work is done pumping ll of the wter over the top edge of the tnk? (b) How much work is done pumping ll of the wter to point 3 m bove the top of the tnk?. A sphericl tnk with rdius 4 m is full of wter. How much work is done lifting ll of the wter to the top of the tnk? () Which requires the most work to empty? Justify your response with detiled explntion. (b) Which requires the lest work to empty?

43 5.4 more work All of the continers shown below hve the sme totl height nd t ech height x bove the ground they ll hve the sme cross-sectionl re.. A spring requires force of 5x dyn to keep it compressed x cm from its nturl length. How much work is done compressing the spring: () 7 cm from its nturl length? (b) cm from its nturl length?. The figure below shows the force required to keep spring tht does not obey Hooke s Lw stretched beyond its nturl length of 3 cm. About how much work is done stretching it: () from length of 3 cm to length 33 cm? (b) from length of 8 cm to length 33 cm? () Which requires the most work to empty? Justify your response with detiled explntion. (b) Which requires the lest work to empty? 7. The figure below shows the force required to move box long rough surfce. How much work is done pushing the box: () from x = to x = 5 feet? (b) from x = 3 to x = 5 feet? 8. How much work is done pushing the box in the figure bove: () from x = 3 to x = 7 feet? (b) from x = to x = 7 feet? 9. A spring requires force of 6x ounces to keep it stretched x inches pst its nturl length. How much work is done stretching the spring: () from its nturl length (x = ) to 3 inches beyond its nturl length? (b) from its nturl length to 6 inches beyond its nturl length?. Approximtely how much work is done stretching the defective spring in the previous problem: () from length of 3 cm to length 6 cm? (b) from length of 3 cm to length 35 cm? 3. A 3-kg object ttched to spring hung from the ceiling stretches the spring 5 cm. How much work is done stretching the spring 4 more cm? 4. A -lb fish stretches spring 3 in. How much work is done stretching the spring 3 more inches? 5. A pylod of mss kg sits on the surfce of the steroid Ceres, dwrf plnet tht is the lrgest object in the steroid belt between Mrs nd Jupiter. Ceres hs dimeter 95 km nd mss kg. How much work is required to lift the pylod from the steroid s surfce to n ltitude of () km? (b) km? (c) 5 km? 6. Clculte the mount of work required to lift you from the surfce of the Erth s moon to n ltitude of km bove the moon s surfce. (The moon s rdius is pproximtely,737.5 km nd its mss is bout 7.35 kg.)

44 43 pplictions of definite integrls 7. Clculte the mount of work required to lift you from the surfce of the Erth s moon (see previous problem) to n ltitude of: () km. (b) 4 km. (c), km. 8. An object locted t the origin repels you with force inversely proportionl to your distnce from the object (so tht f (x) = k where x is your x distnce from the object, mesured in feet). When you re feet wy from the origin, the repelling force is. pound. How much work must you do to move: () from x = to x =? (b) from x = to x =? (c) from x = to x =.? 9. An object locted t the origin repels you with force inversely proportionl to the squre of your distnce from the object (so tht f (x) = k x where x is your distnce from the object, mesured in meters). When you re m wy from the origin, the repelling force is. N. How much work must you do to move: () from x = to x =? (b) from x = to x =? (c) from x = to x =.? 3. A student sid I ve got work long line shortcut tht lwys seems to work. I figure the verge force nd then multiply by the totl distnce. Will it lwys work? () Will it? Justify your nswer. (Hint: Wht is the formul for verge force? ) (b) Is this shortcut? Work Along Curved Pth If the loction of moving object is defined prmetriclly s x = x(t) nd y = y(t) for t b (where t often represents time), nd the force required to overcome friction t time t is given s f (t), we cn represent the work done moving long the (possibly curved) pth s definite integrl. Prtitioning, b] into n subintervls of the form t k, t k ], we cn choose ny c k in t k, t k ] nd pproximte the force required on t k, t k ] by f (c k ) so tht the work done between t = t k nd t = t k is pproximtely: f (c k ) x k ] + y k ] = f (c k ) ] ] xk yk + t t k t k k The totl work done between times t = nd t = b is then: ] ] xk yk b f (c k ) + t t k t k k n k= f (t) x (t)] + y (t)] dt In 3 35, find the work done s n object is moved long the given prmetric pth (with distnce mesured in meters), where f (t) (in newtons) is the force required t time t (in seconds). If necessry, pproximte the vlue of the integrl using technology. 3. f (t) = t, x(t) = cos(t), y(t) = sin(t), t π 3. f (t) = t, x(t) = t, y(t) = t, t 33. f (t) = t, x(t) = t, y(t) = t, t

45 5.4 more work f (t) = sin(t), x(t) = t, y(t) = 3t, t π: 35. f (t) = t, x(t) = cos(t), y(t) = sin(t), t π: (Cn you find geometric wy to clculte the shded re?) 5.4 Prctice Answers. The work done lifting the person nd the stretcher is: (3 lb + lb) (3 ft) = 4 ft-lbs The work done lifting smll piece of cble with length x ft t n initil height of x feet bove the ground is: ( lb ) ( ) ( x ft) (3 x) ft = (6 x) x ft-lbs ft so the work done lifting the cble is: n (6 x) x k= 3 nd the totl work is = 5 ft-lbs. (6 x) dx = 9 ft-lbs. The work done lifting the person nd the stretcher is: ( (5 kg + 5 kg) 9.8 m ) sec (3 m) = (55 N) (3 m) = J The work done lifting smll piece of cble with length x m t n initil height of x m bove the ground is: ( 3 ) kg ( 9.8 m ) ( ) m sec ( x m) (3 x) m = 3.7(3 x) x J so the work done lifting the cble is: n 3 3.7(3 x) x 3.7 (3 x) dx = 47.5 J k= nd the totl work is = 7658 J.

46 434 pplictions of definite integrls 3. The totl work done is pproximtely ] π(.4) (3.5) + π(.6) (4.5) + π(.5) (5.5) + (.) (6.5) (.5787) or oz-in.35 ft-lbs. 4. We cn use the sme integrl s in the solution to Exmple 4, but insted integrte from y = 3 to y = 6: 6 ( W = 9π y y 3) dy = 9π 3 3 y3 ] 6 4 y4 3 ( = 9π (7 34) 9 8 )] = π erg 4 or bout, 79 erg =.79 J. 5. The totl mount of liquid in the trough is 4 5 = ft3, so we need to lift the top ft 3 of liquid out of the trough. To find the height seprting the bottom ft 3 of liquid from the rest, we cn recll tht (from our similr-tringles computtion), the width t height h is w = h, so the volume of liquid between height y = nd height y = h is: = h h 5 h = 8 h = The work to lift the top ft 3 of liquid is thus: 4 ( 75 6y y ) dy = 75 3y ] 4 3 y3 ( = ) ( 4 6 )] 3 3 or bout, ft-lbs. 6. We cn use the sme integrl s in the solution to Exmple 6, but insted integrte from x = 5 to x = : ] 7 7x dx = x = = 6.5 in-lbs =.875 ft-lbs According to Hooke s Lw, lb = k (8 in 3 in) k = 5, so stretching the spring from 5 3 = in to 3 = 7 in beyond its nturl length requires: 7 ] 7 5 x dx = 5 x = 9 in-lb = 3 4 ft-lb 8. The work required to lift the pylod from x = R to x = 3R is: 3R ] GMm 3R x dx = GMm = GMm x 3R + ] = GMm R 6R R R

47 5.5 volumes: tubes Volumes: Tubes In Section 5., we devised the disk method to find the volume swept out when region is revolved bout line. To find the volume swept out when revolving region bout the x-xis (see mrgin), we mde cuts perpendiculr to the x-xis so tht ech slice ws (pproximtely) disk with volume π (rdius) (thickness). Adding the volumes of these slices together yielded Riemnn sum. Tking limit s the thicknesses of the slices pproched, we obtined definite integrl representtion for the exct volume tht hd the form: b π f (x)] dx The disk method, while useful in mny circumstnces, cn be cumbersome if we wnt to find the volume when region defined by curve of the form y = f (x) is revolved bout the y-xis or some other verticl line. To revolve the region bout the y-xis, the disk method requires tht we rewrite the originl eqution y = f (x) s x = g(y). Sometimes this is esy: if y = 3x then x = y 3. But sometimes it is not esy t ll: if y = x + e x, then we cnnot solve for x s n elementry function of y. Refer to Exmples 6(b) nd 6(c) from Section 5. to refresh your memory. The Tube Method Prtition the x-xis (s we did in the disk method) to cut the region into thin, lmost-rectngulr verticl slices. When we revolve one of these slices bout the y-xis (see below), we cn pproximte the volume of the resulting tube by cutting the wll of the tube nd rolling it out flt: to get thin, solid rectngulr box. The volume of the tube is pproximtely the sme s the volume of the solid box: V tube V box = (length) (height) (thickness) = (π rdius]) (height) ( x k ) ( ) = (πc k ) f (c k ) x k where c k is (s usul) ny point chosen from the intervl x k, x k ].

48 436 pplictions of definite integrls The volume swept out when we revolve the whole region bout the y-xis is (pproximtely) the sum of the volumes of these tubes, which is Riemnn sum tht converges to definite integrl: n ( ) (πc k ) f (c k ) x k k= b πx f (x) dx Exmple. Use definite integrl to represent the volume of the solid generted by rotting the region between the grph of y = sin(x) (for x π) nd the x-xis round the y-xis. Solution. Slicing this region verticlly (see mrgin for representtive slice), yields slices with width x nd height sin(x). Rotting slice locted x units wy from the y-xis results in tube with volume: ( ) π (rdius) (height) (thickness) = π (x) sin(x) x where the rdius of the tube (x) is the distnce from the slice to the y- xis nd the height of the tube is the height of the slice (sin(x)). Adding the volumes of ll such tubes yields Riemnn sum tht converges to definite integrl: π π (rdius) (height) dx = π πx sin(x) dx You ll lern how to find n ntiderivtive for x sin(x) in Section 8. or, for now, you cn look for this pttern in Appendix I. We don t (yet) know how to find n ntiderivtive for x sin(x) but we cn use technology (or numericl method from Section 4.9) to compute the vlue of the integrl, which turns out to be π Prctice. Use definite integrl to compute the volume of the solid generted by rotting the region in the first qudrnt bounded by y = 4x x bout the y-xis. If we hd sliced the region in Exmple horizontlly insted of verticlly, the rotted slices would hve resulted in wshers ; pplying the wsher method from Section 5. yields the integrl: π (π rcsin(y)) (rcsin(y)) ] dy Furthermore, the wsher-method integrl in this sitution is more chllenging to set up thn the integrl using the tube method, so the tube method is the most efficient choice on ll counts. The vlue of this integrl is lso π, but finding n ntiderivtive for this integrnd will be much more chllenging thn finding n ntiderivtive for x sin(x). Rotting About Other Axes The tube method extends esily to solids generted by rotting region bout ny verticl line (not just the y-xis).

49 5.5 volumes: tubes 437 Exmple. Use definite integrl to represent the volume of the solid generted by rotting the region between the grph of y = sin(x) (for x π) nd the x-xis round the line x = 4. Solution. The region is the sme s the one in Exmple, but here we re rotting tht region bout different verticl line: Verticl slices gin generte tubes when rotted bout x = 4; the only difference here is tht the rdius for slice locted x units wy from y-xis is now 4 x (the distnce from the xis of rottion to the slice). The volume integrl becomes: π π (rdius) (height) dx = π π(4 x) sin(x) dx which turns out to be π(8 π) Use technology (or tble of integrls) to verify this numericl result. Prctice. Use definite integrl to compute the volume of the solid generted by rotting the region in the first qudrnt bounded by y = 4x x bout the line x = 7. More Generl Regions The tube method lso extends esily to more generl regions. Volumes of Revolved Regions ( Tube Method ) If then the region constrined by the grphs of y = f (x) nd y = g(x) nd the intervl, b] is revolved bout verticl line x = c tht does not intersect the region the volume of the resulting solid is: Mny textbooks refer to this method s the method of cylindricl shells or the shell method, but cylindricl shells is mouthful (compred with tube ) nd shell method is not precise, s shells re not necessrily cylindricl. V = b π x c f (x) g(x) dx The bsolute vlues pper in the generl formul becuse the rdius nd the height re both distnces, hence both must be positive. Exmple 3. Compute the volume of the solid generted by rotting the region between the grphs of y = x nd y = x for x 4 round the y-xis using () verticl slices nd (b) horizontl slices. You cn ensure tht these ingredients in your tube-method integrl will be positive by lwys subtrcting smller vlues from lrger vlues: think right left for x-vlues nd top bottom for y- vlues.

50 438 pplictions of definite integrls Solution. () Verticl slices (see mrgin) result in tubes when rotted bout the y-xis, nd slice x units wy from the y-xis results in tube of rdius x nd height x x, so the volume of the solid is: 4 ] πx x x dx = π = π 4 x 3 x ] dx = π ( ) ( x4 3 x3 )] = 48π 3 or bout (b) Horizontl slices result in wshers when rotted bout the y-xis, but we hve new problem: the lower slices (where y 4) extend from the line x = on the left to the line y = x on the right, while the upper slices (where 4 y 6) extend from the prbol y = x on the left to the line x = 4 on the right. This requires us to use two integrls to compute the volume: y=4 y= π y ] y=6 dy + π 4 ( y) ] dy y=4 ] 4 Evluting these integrls is strightforwrd, but setting them up ws more timeconsuming thn using the tube method. Evluting these integrls lso results in volume of 48π Prctice 3. Find the volume of the solid formed by rotting the region between the grphs of y = x nd y = x for x 4 round x = 3. Prctice 4. Compute the volume of the solid generted by rotting the region in the first qudrnt bounded by the grphs of y = x, y = x + nd x = 4 round () the y-xis (b) the x-xis. Exmple 4. Compute the volume of the solid swept out by rotting x the region in the first qudrnt between the grphs of y = nd y = x bout the x-xis. Solution. Grphing the region (see mrgin), it is pprent tht the curves intersect where: x = x x = x x = Slicing the region verticlly results in two cses: when x, the slice extends from the x-xis to the curve y = x ; when x, the Both types of slices re perpendiculr to the x-xis, so the width of ech slice is of the form x nd our integrls should involve dx. slice extends from y = x to y = x. Rotting the first type of slice bout the x-xis results in disk; rotting the second type of slice bout the x-xis results in wsher. Using the disk method for the first intervl nd the wsher method for the second intervl, the volume of the solid is: π ] x dx + π ( ) ] x ( ) x dx

51 5.5 volumes: tubes 439 Evluting these integrls is strightforwrd: x x ] π dx + π (x ) dx = π x 4 ] + π ] x x = π 4 If you hd insted sliced the region horizontlly, you would only need one type of slice (see mrgin). Rotting horizontl slice round the x-xis results in tube. Becuse this slice is perpendiculr to the y-xis, the thickness of the slice is of the form y, so the tube-method integrl will include dy nd we will need to formulte the rdius nd height of the tube in terms of y. The rdius of the slice is merely y, the distnce between the slice nd the x-xis. The height of the slice is its length, which is the distnce between the two curves. The left-hnd curve is: x y = y = x x = y nd the right-hnd curve is: y = x y = x x = y + so the distnce between the two curves is: ( ) ( y + y ) = y The curves intersect where: y + = y y = y = ±; from the grph we cn see tht the bottom of the region corresponds to y = nd the top of the region is t y =. Applying the tube method, the volume of the solid is: y= y= πy y ] dy = π y y 3] ] y dy = π y4 = π 4 which grees with the result bove from the disk+wsher method. This ppliction of the tube method rottes horizontl slice round horizontl xis; in previous tube-method pplictions we hve only rotted verticl slice bout verticl xis. Either option results in tube, nd the generl formul on pge 433 cn be further extended to this new sitution s we hve done here by swpping the roles of x nd y, Prctice 5. Compute the volume of the solid swept out by rotting x the region in the first qudrnt between the grphs of y = nd y = x bout () the line x = 5 (b) the line y = 5. Which Method Is Best? In theory, both the wsher method nd the tube method will work for ny volume-of-revolution problem involving horizontl or verticl xis. In prctice, however, one of these methods is usully esier to use thn the other but which one is esier depends on the prticulr region nd type of xis. As we hve seen, chllenges my include: We will investigte method for computing volumes of solids formed by rotting region round tilted xes in Section 5.6. The necessity to split the region into two (or more) pieces, resulting in two (or more) integrls.

52 44 pplictions of definite integrls The difficulty (or impossibility) of solving n eqution of the form y = f (x) for x or n eqution of the form x = g(y) for y. The difficulty (or impossibility) of finding n ntiderivtive for the resulting integrnd. With experience (nd lots of prctice) you will begin to develop n intuition for which method might be the best choice for prticulr sitution. Sketching the region long with representtive horizontl nd verticl slices is vitl first step. The method tht voids the need to split the region up into more thn one piece is often but not lwys the superior choice. Avoiding the need to find n inverse function for boundry curve should lso be priority. Finlly, if you need n exct vlue nd one method results in chllenging ntiderivtive serch, strt over nd try the other method. 5.5 Problems In Problems 6, sketch the region nd clculte the volume swept out when the region is revolved bout the specified verticl line.. The region in the first qudrnt between the curve y = x nd the x-xis is rotted bout the y-xis.. The region in the first qudrnt between the curve y = x x nd the x-xis is rotted bout the y-xis. 3. The region in the first qudrnt between between y = x, y = x nd the line x = 3 is rotted bout the line x = The region in the first qudrnt between the curve y =, the x-xis nd the line x = 3 is rotted bout the + x y-xis. 7. The region in the first qudrnt between the grphs of y = ln(x), y = x nd x = The region in the first qudrnt between the grphs of y = e x, y = x nd x =. 9. The region between y = x nd y = 6 x for x 4.. The shded region in the figure below.. The shded region in the figure below. 5. The region between y = x, y = nd x = is 3 rotted bout the line x = The region between y = x, y = x, x = nd x = 3 is rotted bout the line x =. In Problems 7, use definite integrl to represent the volume swept out when the given region is revolved bout the y-xis, then use technology to evlute the integrl.

53 5.5 volumes: tubes Problems In Problems 3, set up n integrl to clculte the volume swept out when the region between the given curves is rotted bout the specified xis, using ny pproprite method (disks, wshers, tubes). If possible, work out n exct vlue of the integrl; otherwise, use technology to find n pproximte numericl vlue.. y = x, y = x 4, bout the y-xis 3. y = x, y = x 4, bout the y-xis 4. y = x, y = x 4, bout the x-xis 5. y = sin(x ), y =, x =, x = π, bout x = 6. y = cos(x ), y =, x =, x = π, bout x = 7. y =, y =, x =, x = x, bout x = 8. y = x, y =, x =, x =, bout y = 9. y = x, y = x 4, bout x = 3. y = x, y = x 4, bout y = 3. y = x, y = x 4, bout y = 3. y = x, y = x 4, bout x = 3 3. y =, y =, x =, x = bout x = + x 4. y = + x, y =, x =, x = 3, bout x = 5. y =, y =, x =, bout x = + x 6. y = + x, y =, bout x = 7. y = x, y = x, y =, x = 3, bout x = 4 8. y = x, y = x, y =, x = 3, bout x = 4 9. y = x, y = x, y =, x = 3, bout y = 4 3. y = x, y = x, y =, x = 3, bout y = Prctice Answers. Grph the region (see mrgin) nd note tht the curve y = 4x x intersects the x-xis where 4x x = x(4 x) = x = or x = 4. Rotting verticl slice round the y-xis results in tube with rdius x (the distnce between the slice nd the y-xis) nd height 4x x so the volume of the solid is: 4 ( πx 4x x ) dx = π = π 4 4x x 3] dx 4 3 x3 4 x4 ] 4 = 8π The region here is identicl to the region in Prctice, but we re now rotting slice round the xis x = 7, so the rdius of the resulting tube is x ( 7) = x + 7 (the distnce from the slice t loction x to the xis of rottion). The volume of the solid is therefore: 4 π(x + 7) ( 4x x ) dx = π = π 4 8x 3x x 3] dx 4x x 3 4 x4 ] 4 = 9π 63

54 44 pplictions of definite integrls 3. Rotting verticl slice (see mrgin figure) bout the line x = 3 results in tube with rdius 3 x nd height x x, so the volume of the solid is: 4 π(3 x)(x x) dx = π 4 = π 3 x x3 ] 4 4 x4 = π 3x + 4x x 3] dx 39 3 ] = 764π Grph the region (see mrgin) nd drw representtive verticl slice. (Horizontl slices would require splitting the region into two pieces why?) () Rotting the verticl slice bout the y-xis results in tube of rdius x (the distnce from the slice to the y-xis) nd height (x + ) x, nd the region sits between x = nd x = 4 so the volume of the solid is: 4 πx x + x ] dx = π = π 4 ] x + x x 3 dx 3 x3 + x 5 x 5 ] 4 = 496π 5 4 (b) Rotting the verticl slice round the x-xis results in wsher with big rdius x + (the distnce from the x-xis to the curve frthest from the x-xis) nd smll rdius x (the distnce from the x-xis to the closer curve) so the volume of the solid is: 4 π (x + ) ( x ) ] dx = π 4 ] x + x + dx = π This region is the sme s the one in Exmple 4, where it ws pprent tht slicing horizontlly resulted in single type of slice (compred with verticl slices, which required us to split the region into two pieces). () Rotting horizontl slice round the verticl line x = 5 results in wshers with thickness y (so our integrl will involve dy), big rdius 5 y (the distnce between the xis of rottion nd the frthest curve) nd smll rdius 5 ( y + ) = 4 y (the distnce between the xis of rottion nd closest curve). Applying the wsher method, the volume of the solid is: ( π 5 y ) ( 4 y ) ] dy = 4π (b) Rotting horizontl slice round the horizontl line y = 5 results in tube of rdius 5 y (the distnce between the slice nd the xis of rottion) nd height y (the length of the slice). Applying the tube method, the volume of the solid is: ( π(5 y) y ) dy = 37π 9.4 6

55 5.6 moments nd centers of mss Moments nd Centers of Mss This section develops method for finding the center of mss of thin, flt shpe the point t which the shpe will blnce without tilting (see mrgin). Centers of mss re importnt becuse in mny pplied situtions n object behves s though its entire mss is locted t its center of mss. For exmple, the work required to pump the wter in tnk to higher point is the sme s the work required to move smll object with the sme mss locted t the tnk s center of mss to the higher point (see mrgin), much esier problem (if we know the mss nd the center of mss of the wter). Volumes nd surfce res of solids of revolution cn lso become esy to clculte if we know the center of mss of the region being revolved. Point-Msses in One Dimension Before investigting the centers of mss of complicted regions, we consider point-msses (nd systems of point-msses), first in one dimension nd then in two dimensions. Two people with different msses cn position themselves on seesw so tht the seesw blnces (see mrgin). The person on the right cuses the seesw to wnt to turn clockwise bout the fulcrum, nd the person on the left cuses it to wnt to turn counterclockwise. If these two tendencies re equl, the seesw will blnce on the fulcrum. A mesure of this tendency to turn bout the fulcrum is clled the moment bout the fulcrum of the system, nd its mgnitude is the product of the mss nd the distnce from the mss to the fulcrum. In generl, the moment bout the origin, M, produced by mss m t loction x is m x, the product of the mss nd the signed distnce of the point-mss from the origin (see mrgin). For system of n msses m, m,..., m n t loctions x, x,..., x n, respectively, the totl mss of the system is: In this seesw exmple, we need to imgine tht the seesw is constructed using very lightweight yet sturdy substnce, so tht its mss is negligible compred with the msses of the two people. m = m + m + + m n = n m k k= nd the moment bout the origin of the system is: M = m x + m x + + m n x n = n m k x k k= If the moment bout the origin is positive, then the system tends to rotte clockwise bout the origin. If the moment bout the origin is negtive, then the system tends to rotte counterclockwise bout the origin. If the moment bout the origin is zero, then the system does not tend to rotte in either direction bout the origin: it blnces on fulcrum locted t the origin.

56 444 pplictions of definite integrls You hve seen this br nottion before, in conjunction with the verge vlue of function. Here we cn think of x s weighted verge. We cn fctor x out of the second sum becuse it is constnt. The moment bout point x = p, M p, produced by mss m t loction x = x is the product of the mss nd the signed distnce of x from the point p: m (x p). The moment bout point x = p produced by msses m, m,..., m n t loctions x, x,..., x n, respectively, is: M p = m (x p) + m (x p) + + m n (x n p) = n m k (x k p) k= The point t which system of point-msses blnces is clled the center of mss of the system, written x (pronounced x-br ). Becuse the system blnces t x = x, the moment bout x, M x, must be. Using this fct (nd summtion properties), we obtin formul for x: n n ] n ] = M x = m k (x k x) = m k x k m k x k= k= = n k= m k x k ] x n k= m k ] = M x m so x m = M nd solving for x yields the following formul. k= The center of mss of system of point-msses m, m,..., m n t loctions x, x,..., x n is: x = M m = n k= m k x k n k= m k k m k x k A single point-mss with mss m (the totl mss of the system) locted t x (the center of mss of the system) produces the sme moment bout ny point on the line s the whole system: M p = n k= = m m k (x k p) = ( ) M m p n k= = m (x p) m k x k ] p n k= m k ] = M pm For mny purposes, we cn think of the mss of the entire system s being concentrted t x. Exmple. Find the center of mss of the system consisting of the first three point-msses listed in the mrgin tble. Solution. m = = 6 nd M = ()( 3) + (3)(4) + ()(6) = so: x = M m = 6 = The system of three point-msses will blnce on fulcrum t x =. Prctice. Find the center of mss of the system consisting of the lst three point-msses listed in the mrgin tble.

57 5.6 moments nd centers of mss 445 Point-Msses in Two Dimensions The concepts of moments nd centers of mss extend nicely from one dimension to system of msses locted t points in plne. For knife edge fulcrum locted long the y-xis (see mrgin), the moment of point-mss with mss m locted t the point (x, y ) is the product of the mss nd the signed distnce of the point-mss from the y-xis: m x. This tendency to rotte bout the y-xis is clled the moment bout the y-xis, written M y. Here, M y = m x. Similrly, pointmss with mss m locted t the point (x, y ) hs moment bout the x-xis (see mrgin): M x = m y. For system of msses m k locted t the points (x k, y k ), the totl mss of the system is (s before): As in the seesw exmple, we need to imgine tht the point-msses re sitting on thin yet strong plte of negligible mss compred with the pointmsses. m = m + m + + m n = while the moment bout the y-xis is: n m k k= M y = m x + m x + + m n x n = nd the moment bout the x-xis is: M x = m y + m y + + y n x n = n m k x k k= n m k y k k= At first, it my seem confusing tht the formul for M y would involve x nd the formul for M x would involve y, but keep in mind tht n eqution for the y-xis is x =, so we could write the moment bout the y-xis s M x= nd the moment bout the x-xis s M y=. The center of mss of this two-dimensionl system is point (x, y) such tht ny line tht psses through this point is blncing fulcrum for the system. So we need the moment bout ny such line including x = x nd y = y to be zero: = M x=x = n n ] n ] m k (x k x) = m k x k x m k = M y xm k= k= k= If we cn find such point, then the system will blnce on single pointfulcrum locted t the center of mss. so x = M y m, nd similr rithmetic shows tht y = M x m. The center of mss of system of point-msses m, m,..., m n t loctions (x, y ), (x, y ),..., (x n, y n ) is the point (x, y) where: x = M y m = n k= m k x k n k= m k nd y = M x m = n k= m k y k n k= m k

58 446 pplictions of definite integrls The rithmetic needed to prove this sttement is similr to rithmetic we did to prove the corresponding ssertion for one-dimensionl system. k m k x k y k A single point-mss with mss m (the totl mss of the system) locted t (x, y) (the center of mss of the system) produces the sme moment bout ny line s the whole system does bout tht line. For mny purposes, we cn think of the mss of the entire system being concentrted t (x, y). Exmple. Find the center of mss of the system consisting of the first three point-msses listed in the mrgin tble. Solution. m = = 6 nd M y = ()( 3) + (3)(4) + ()(6) = while M x = ()(4) + (3)( 7) + ()( ) = 5 so: x = M y m = 6 = nd y = M x m = 5 6 =.5 The system of three point-msses will blnce on ny fulcrum pssing through the point (,.5). Prctice. Find the center of mss of the system consisting of ll five point-msses listed in the mrgin tble. Centroid of Region When we move from discrete point-msses to continuous regions in plne, we move from finite sums nd rithmetic to limits of Riemnn sums, definite integrls nd clculus. The following discussion extends ides nd clcultions from point-msses to uniformly thin, flt pltes (clled lmin) tht hve uniform density throughout (given s mss per re, such s grms per cm ). The center of mss of one of these pltes is the point (x, y) t which the plte blnces without tilting. It turns out tht for pltes with uniform density, the center of mss (x, y) depends only on the shpe (nd loction) of the region of the plne covered by the plte nd not on the (constnt) density. In these uniform-density situtions, we cll the center of mss the centroid of the region. Throughout the following discussion, you should notice tht ech finite sum tht ppered in the discussion of point-msses hs n integrl counterprt for these thin pltes. Rectngles The components of Riemnn sum typiclly involve res of rectngles, so it should come s no surprise tht the bsic shpe used to extend point-mss concepts to regions is the rectngle. The totl mss of rectngulr plte is the product of the re of the plte nd its (constnt) density: m = mss = (re) (density). We will ssume tht the center of mss of thin, rectngulr plte is locted hlfwy up nd hlfwy cross the rectngle, t the point where the digonls of the rectngle cross (see mrgin).

59 5.6 moments nd centers of mss 447 The moments of the rectngle bout n xis cn be found by treting the rectngle s single point-mss with mss m locted t the center of mss of the rectngle. Exmple 3. Find the moments bout the x-xis, y-xis nd the line x = 5 of the thin, rectngulr plte shown in the mrgin. Solution. The density of the plte is 3 g/cm nd the re of the plte is ( cm) (4 cm) = 8 cm so the totl mss is: m = ( 8 cm ) ( 3 g ) cm = 4 g The center of mss of the rectngulr plte is (x, y) = (3, 4). The moment bout the x-xis is the product of the mss nd the signed distnce of the mss from the x-xis: M x = (4 g) (4 cm) = 96 g-cm. Similrly, M y = (4 g) (3 cm) = 7 g-cm. The moment bout the line x = 5 is M x=5 = (4 g) (5 3] cm) = 48 g-cm. To find the moments nd center of mss of plte mde up of severl rectngulr regions, we cn simply tret ech of the rectngulr pieces s point-mss concentrted t its center of mss, then tret the plte s system of discrete point-msses. Exmple 4. Find the centroid of the region in the mrgin figure. Solution. We cn divide the plte into two rectngulr pltes, one with mss 4 g nd center of mss (, 4), nd the other with mss g nd center of mss (3, 3). The totl mss of the pir of point-msses is m = 4 + = 36 g, nd the moments bout the xes re M x = (4 g) (4 cm) + ( g) (3 cm) = 3 g-cm nd M y = (4 g) ( cm) + ( g) (3 cm) = 6 g-cm. So: x = M y m = 6 g-cm 36 g = 5 3 cm nd y = M x m ( The centroid of the plte is locted t 53, 3 3 g-cm = = 36 g 3 cm ). Prctice 3. Find the centroid of the region in the mrgin figure. To find the center of mss of thin, non-rectngulr plte, we will slice the plte into nrrow, lmost-rectngulr pltes nd tret the collection of lmost-rectngulr pltes s system of point-msses locted t the centers of mss of the lmost-rectngles. The totl mss nd moments bout the xes for the system of point-msses will be Riemnn sums. By tking limits s the widths of the lmost-rectngles pproch, we will obtin exct vlues for the mss nd moments s definite integrls

60 448 pplictions of definite integrls x for Region The Greek letter ρ (pronounced row, s in row your bot ) is often used to represent the density of region. Suppose f (x) g(x) on the intervl, b] nd R is plte of uniform density (= ρ) sitting on the region between the grphs of f (x) nd g(x) nd the lines x = nd x = b (see mrgin figure). If we prtition the intervl, b] into n subintervls of the form x k, x k ] nd choose the points c k to be the midpoints of these subintervls, then the slice between verticl cuts t x = x k nd x = x k is pproximtely rectngulr nd hs mss pproximtely equl to: (re) (density) = (height) (width) (density) f (c k ) g (c k )] (x k x k ) ρ = ρ f (c k ) g (c k )] x k So the mss of the whole plte is pproximtely m = n ρ f (c k ) g (c k )] x k k= b ρ f (x) g(x)] dx = ρ A where A is the re of the region R. The moment bout the y-xis of ech lmost-rectngulr slice is the product of the mss of the slice (m) nd the distnce from the centroid of the lmost-rectngle to the y-xis. The x-coordinte of tht centroid is locted t x = c k, so the distnce from the centroid to the y-xis is c k = c k. The moment of the lmost-rectngle bout the y-xis is therefore: m k c k = (ρ f (c k ) g (c k )] x k ) c k so the moment of the entire plte bout the y-xis is (pproximtely): M y = n ρc k f (c k ) g (c k )] x k k= b The x-coordinte of the centroid of the plte is therefore: x = M y m = ρ b b x f (x) g(x)] dx ρ b f (x) g(x)] = dx b ρx f (x) g(x)] dx x f (x) g(x)] dx f (x) g(x)] dx The density constnt ρ is fctor of both M y nd m, so it cncels nd hs no effect on the vlue of x. The vlue of x depends only on the shpe nd loction of the region R. If the bottom boundry of R is the x-xis, then g(x) = nd the previous formuls simplify to: b b m = ρ f (x) dx, M y = ρ x f (x) dx nd x = M b y m = b x f (x) dx f (x) dx Prctice 4. Find the x-coordinte of the centroid of the region between f (x) = x, the x-xis nd x =.

61 5.6 moments nd centers of mss 449 y for Region To find y, the y-coordinte of the centroid of R, we need to find M x, the moment of R bout the x-xis. For verticl prtitions of R (see mrgin), the moment of ech nrrow strip bout the x-xis, M x, is the product of the strip s mss nd the signed distnce between the centroid of the strip nd the x-xis. We ve lredy computed the mss: m k = ρ f (c k ) g (c k )] x k Becuse ech strip is nerly rectngulr, the centroid of the k-th strip is roughly hlfwy up the strip, t point midwy between f (c k ) nd g (c k ), so we cn verge those function vlues to compute: y k f (c k) + g (c k ) The moment bout the x-xis for this strip is thus: ] f (ck ) + g (c ρ f (c k ) g (c k )] x k k ) = ρ ( f (c k )) (g (c k )) ] x k Adding up the moments of ll n strips yields: M x = y = M x m n k= ρ ( f (c k )) (g (c k )) ] b ρ x k ( f (x)) (g (x)) ] dx The y-coordinte of the centroid of the plte is therefore: ( f (x)) (g (x)) ] dx = ρ b ρ b f (x) g(x)] dx = b ( f (x)) (g (x)) ] dx b f (x) g(x)] dx If the bottom boundry of R is the x-xis, then g(x) = nd the previous formuls simplify to: M x = ρ b x f (x)] dx nd y = M x m = b f (x)] dx b f (x) dx Exmple 5. Find the y-coordinte of the centroid of the region R bounded below by the x-xis nd bove by the top hlf of circle of rdius r centered t the origin (see mrgin). Solution. An eqution for the circle is x + y = r so the top hlf is given by f (x) = y = r x, nd g(x) =. The mss of the region is: m = r r r ρ r x dx = ρ r x dx = ρ re of R] = ρ πr r The moment of R bout the y-xis is: r M y = ρx r x = ρ ( r x ) 3 3 so x =. r ] x=r x= r = Could you hve guessed this result merely by looking t the region?

62 45 pplictions of definite integrls The moment of R bout the x-xis is: M x = r = ρ r ρ r x ] dx = ρ r r r x ] r 3 x3 = ρ r 4 3 r3 = ρ 3 r3 r x ] dx Could you hve guessed tht centroid would be locted bit less thn hlfwy bove the bottom edge of the semicircle, merely by looking t the region? so y = ρ 3 r3 ρπ r = 4 3π r.444r. Prctice 5. Show tht the centroid of tringulr ( region with vertices b (, ), (, h) nd (b, ) is locted t (x, y) = 3, h ). 3 The following tble summrizes nd compres formuls for computing moments nd centers of mss for system of point-msses in plne (using sums) nd for region in plne (using integrls). The integrl formuls pper in form for clculting moments of region R bounded by the grphs of two functions, f (x) nd g(x), nd two verticl lines, x = nd x = b, where f (x) g(x) for x b. totl mss: m = moment bout y-xis (x = ): M y = moment bout x-xis (y = ): M x = center of mss (ρ constnt): point-msses in plne n k= n k= m k m = m k x k M y = n m k y k M x = k= x = M y m, y = M x m region R between f nd g b b b ρ f (x) g(x)] dx = ρ Are (R) x = M y m, y = M x m ρx f (x) g(x)] dx ρ ( f (x)) (g(x)) ] dx With the knowledge of Riemnn sums you hve developed, you should be ble to set up integrls to compute msses nd moments for regions bounded by curves of the form x = g(y), nd del with situtions where the density of thin plte is function of x or y. While the integrl formuls bove re often useful, it is importnt tht you understnd the process used to obtin these formuls in order to compute moments nd centroids of more generl regions. Exmple 6. Find the centroid of the region R bounded by the grphs of y = x nd y = x 3. Solution. The curves intersect where x = x 3 x x 3 = x ( x) = x = or x =. A grph (see mrgin) helps confirm tht x x 3 on, ]. If the density of R is ρ then the mss of R is: m = ρ x x 3] dx = ρ 3 x3 ] 4 x4 = ρ The moment of R bout the y-xis is: M y = ρ x x x 3] dx = ρ x 3 x 4] dx = ρ 4 x4 ] 5 x5 = ρ

63 5.6 moments nd centers of mss 45 And the moment of R bout the x-xis is: M x = ρ ( x ) ( x 3) ] dx = ρ 5 x5 ] 7 x7 = ρ 35 so x = M ρ y m = ρ = 3 5 nd y = M ρ x m = 35 ρ =. Plotting the point 35 ( ) 35, 35 (.6,.34) long with R confirms tht it sits inside R (just brely) nd ppers to be resonble cndidte for the centroid. Symmetry Symmetry is very powerful geometric concept tht cn simplify mny mthemticl nd physicl problems, including the tsk of finding centroids of regions. For some regions, we cn use symmetry lone to determine the centroid. Geometriclly, region R is symmetric bout line L if, when R is folded long L, ech point of R on one side of the fold mtches up with exctly one point of R on the other side of the fold (see mrgin). Exmple 7. Sketch two lines of symmetry for ech region shown in the mrgin figure. Solution. See solution to Prctice 6. A very useful fct bout symmetric regions is tht the centroid (x, y) of symmetric region must lie on every line of symmetry of the region. If region hs two different lines of symmetry, then the centroid must lie on ech of them, so the centroid must be locted t the point where the lines of symmetry intersect. Prctice 6. Locte the centroid of ech region in Exmple 7. In Exmple 5, the hlf-disk ws symmetric with respect to the y-xis, so we could hve voided setting up nd evluting the M y integrl by noticing tht (x, y) must be locted on the y-xis (the line x = ) nd concluding tht x =. Work In uniform grvittionl field, the center of grvity of n object is locted t the sme point s its center of mss, nd the work done to lift n object is the product of the object s weight nd the distnce tht the center of grvity of the object is rised: work = (object s weight) (distnce object s center of grvity is rised) In the high jump, this explins the effectiveness of the Fosbury Flop, technique where the jumper ssumes n inverted U position while going over the br (see mrgin): the jumper s body goes over the br while the jumper s center of grvity goes under it, llowing the jumper to cler higher br with no dditionl upwrd thrust. If you know the center of grvity of n object being lifted, some work problems become much esier.

64 45 pplictions of definite integrls We ve lredy solved this problem (s Exmple 5 in Section 5.4) but here we try new pproch using centroids. Exmple 8. The trough shown in the mrgin is filled with liquid weighing 7 pounds per cubic foot. How much work is done pumping the liquid over the wll next to the trough? Solution. This is 3-D problem, but symmetry tells us the centroid of the liquid must be t point.5 feet from either end of the trough, nd foot wy from the wll. The verticl coordinte of the centroid will be the sme s the centroid of the trough s tringulr end region. Using the result of Prctice 5, we cn conclude tht the centroid of the tringle is t height of 3 4 = 8 3. The weight of the liquid is: ( (density) (volume) = 7 lb ) ft 3 (5 ft) ( ft) (4 ft) = 4 lbs nd the distnce the center of grvity must be moved is = 3 ft so the totl work required is: ( ) (4 lbs) 3 ft = 4 3 ft-lbs ft-lbs which grees with the nswer obtined in Section 5.4. Pppus, the lst of the gret Greek geometers, flourished during the first hlf of the fourth century. Theorems of Pppus Two theorems due to Pppus of Alexndri cn mke some volume nd surfce re clcultions reltively esy. Touching the boundry is OK. Theorem of Pppus: Volume of Revolution If plne region R with re A nd centroid (x, y) is revolved round line L in the plne tht does not pss through R then the volume swept out by one revolution of R is the product of A nd the distnce trveled by the centroid. The distnce from the centroid to the line will be the rdius of the circle swept out by the centroid, so the distnce trveled by the centroid is π times this rdius. When L is the x-xis, the volume of the solid is A πy; when L is the y-xis, the volume of the solid is A πx. Exmple 9. Find the volume swept out when the region R bounded by the grphs of y = x nd y = x 3 is revolved round the line x =. Solution. ( ) From Exmple 6, we know the re of R is nd its centroid is 35, 35. The distnce from this point to the line x = is 3 5 = 7 5, so the distnce trveled by the centroid is π 7 5 = 4π. The volume 5 of the solid of revolution is therefore 4π = 7π 5 3.

65 5.6 moments nd centers of mss 453 Theorem. of Pppus: Surfce Are of Revolution If plne region R with perimeter P nd centroid (x, y) is revolved round line L in the plne tht does not pss through R then the surfce re swept out by one revolution of R is the product of P nd the distnce trveled by the centroid. Touching the boundry is OK. When L is the x-xis, the surfce re of the solid is P πy; when L is the y-xis, the surfce re is P πx. Exmple. Find the surfce re of the solid swept out when the squre region R with vertices t (, ), (, ), (, ) nd (, ) is revolved round the line y = 3. Solution. By symmetry, the centroid of the squre is (, ) nd its distnce from y = 3 is 3. The perimeter of the squre is 4, so the surfce re of the solid of revolution is 4 π 3 = 4π Problems. () Find the totl mss nd the center of mss for system consisting of the three point-msses in the tble below left. (b) Where should you locte new object with mss 8 so the new system hs its center of mss t x = 5? (c) Wht mss should you put t x = so the originl system plus the new mss hs its center of mss t x = 6? m k 5 5 x k 4 6 m k x k () Find the totl mss nd the center of mss for system consisting of the four point-msses in the tble bove right. (b) Where should you locte new object with mss so the new system hs its center of mss t x = 6? (c) Wht mss should you put t x = 4 so the originl system plus the new mss hs its center of mss t x = 6? 3. () Find the totl mss nd the center of mss for system consisting of the three point-msses in the tble below. (b) Where should you locte new object with mss so the new system hs its center of mss t (5, )? m k 5 5 x k 4 6 y k () Find the totl mss nd the center of mss for system consisting of the four point-msses in the tble below. (b) Where should you locte new object with mss so the new system hs its center of mss t (3, 5)? m k x k y k 4 7 8

66 454 pplictions of definite integrls In Problems 5, divide the plte shown into rectngles nd semicircles, clculte the mss, moments nd centers of mss of ech piece, then find the center of mss of the plte. Assume the density of the plte is ρ =. Plot the loction of the center of mss for ech shpe. (Refer to Exmple 5 for centroids of semicirculr regions.) 5. Use the figure below left. 6. Use the figure bove right. 7. Use the figure below left. 8. Use the figure bove right. 9. Use the figure below left. 3. y = x, y = y = sin(x), the x-xis, the y-xis, x = π. 5. y = 4 x nd the x-xis for x. 6. y = x, y = x for x =. 7. y = 9 x, y = 3, x =, x = y = x, the x-xis, x =, x =. 9. y = x, the x-xis, x = 9.. y = ln(x), the x-xis, x = e.. y = e x, y = e nd the y-xis.. y = x nd y = x. 3. An empty box in the shpe of cube mesuring foot on ech side weighs pounds. By symmetry, we know its center of mss is 6 inches bove its bottom. When the box is full of liquid with density 6 lb/ft 3, the center of mss of the box-liquid system is gin (due to symmetry) 6 inches bove the bottom of the box. () Find the height of the center of mss of the box-liquid system s function of h, the height of wter in the box. (b) To wht height should you fill the box so tht the box-liquid system hs the lowest center of grvity (nd the gretest stbility)? 4. The empty glss shown below left hs mss of g when empty. Find the height of the center of mss of the glss-wter system s function of the height of wter in the glss.. Use the figure bove right. In Problems 6, sketch the region bounded by the the given curves nd find the centroid of ech region (use technology to evlute integrls, if necessry). Plot the loction of the centroid on your sketch of the region.. y = x, the x-xis, x = 3.. y = x, the x-xis, x =, x =. 5. The empty sod cn shown bove right hs mss of 5 g when empty nd 4 g when full of sod. Find the height of the center of mss of the cn-sod system s function of the height of the sod in the cn.

67 5.6 moments nd centers of mss Give prcticl set of directions someone could ctully use to find the height of the center of grvity of their body with their rms t their sides. How will the height of the center of grvity chnge if they lift their rms? 7. Try the following experiment. Stnd stright with your bck nd heels ginst wll. Slowly rise one leg, keeping it stright, in front of you. Wht hppened? Why? 8. Why cn t two dncers stnd in the position shown below? 9. If shpe hs exctly two lines of symmetry, the lines cn meet t right ngles. Must they meet t right ngles? 3. Sketch regions with exctly two lines of symmetry, exctly three lines of symmetry, nd exctly four lines of symmetry. 3. A rectngulr box is filled to depth of 4 feet with 3 pounds of wter. How much work is done pumping the wter to point feet bove the bottom of the box? 3. A cylinder is filled to depth of feet with 4 pounds of wter. How much work is done pumping the wter to point 7 feet bove the bottom of the cylinder? 33. A sphere of rdius m is filled with wter. How much work is done pumping the wter to point 3 m bove the top of the sphere? 34. A sphere of rdius feet is filled with wter. How much work is done pumping the wter to point 5 feet bove the top of the sphere? 35. The center of squre region with sides of length cm is locted t the point (3, 4). Find the volume swept out when the squre region is rotted: () bout the x-xis. (b) bout the y-xis. (c) bout the line y = 6 (d) bout the line x = 6 (e) bout the line x + 3y = The lower left corner of rectngulr region with n 8-inch bse nd 4-inch height is locted t the point (3, 5). Find the volume swept out when the rectngulr region is rotted: () bout the x-xis. (b) bout the y-xis. (c) bout the line y = x The center of squre region with sides of length cm is locted t the point (3, 4). Find the surfce re swept out when the squre region is rotted: () bout the x-xis. (b) bout the y-xis. (c) bout the line y = 6 (d) bout the line x = 6 (e) bout the line x + 3y = The lower left corner of rectngulr region with n 8-inch bse nd 4-inch height is locted t the point (3, 5). Find the surfce re swept out when the rectngulr region is rotted: () bout the x-xis. (b) bout the y-xis. (c) bout the line y = x Find the volume nd surfce re swept out when the region inside the circle (x 3) + (y 5) = 4 is rotted: () bout the x-xis. (b) bout the y-xis. (c) bout the line y = 9 (d) bout the line x = 6 (e) bout the line x + 3y = 6

68 456 pplictions of definite integrls 4. Find the volume nd surfce re swept out when the center of circle with rdius r nd center (R, ) is rotted bout the x-xis (see below). 4. Find the volumes nd surfce res swept out when the rectngles shown below re rotted bout the line L. (Mesurements re in feet.) Physiclly Approximting Centroids of Regions You cn pproximte the loction of centroid of region experimentlly, even if the region such s stte or country is not described by formul. Cut the shpe out of piece of some uniformly thick mteril, such s pper or crdbord, nd pin n edge to wll. The shpe will pivot bout the pin until its center of mss is directly below the pin (see mrgin) so the center of mss of the shpe must lie directly below the pin, on the line connecting the pin with the center of mss of Erth. Repet the process using different point ner the edge of the shpe to find different line. The center of mss lso lies on the new line, so you cn conclude tht the centroid of the shpe is locted where the two lines intersect (see mrgin). It is good ide to pick third point ner the edge nd plot third line to check tht this third line lso psses through the point of intersection of the first two lines. You cn experimentlly pproximte the popultion center of region by ttching msses proportionl to the popultions of the cities nd then repeting the pin process with this weighted model. The point on the new model where the lines intersect is the pproximte popultion center of the region. 4. Determine the centroid of your stte. 43. Which stte would result in the esiest centroid problem? The most difficult centroid problem?

69 5.6 moments nd centers of mss Prctice Answers. m = = 9; M = ()(6) + (5)( ) + (3)(4) = 8; x = M m = 8 9 ; the three point-msses will blnce on fulcrum locted t x = m = = 4 M y = ()( 3) + (3)(4) + ()(6) + (5)( ) + (3)(4) = 4 M x = ()(4) + (3)( 7) + ()( ) + (5)() + (3)( 6) = 8 x = M y m = 4 4 = nd y = M x m = 8 4 = The five point-msses blnce t the point (, ). 3. There re severl wys to brek the region into esy pieces one wy is to consider the four cm-by- cm squres. The center of mss of ech squre is locted t the center of the squre (t (, ), (4, ), (6, ) nd (4, 4)), nd ech squre hs mss ( 4 cm ) ( 5 g ) cm = g so: m = 4 ( g) = 8 g, M y = () + 4() + 6() + 4() = 3 g-cm nd M x = () + () + () + 4() = g-cm. Therefore x = M y 3 g-cm = = 4 cm nd y = M x m 8 g m.5 cm so the center of mss is locted t (4,.5). 4. For simplicity, let ρ =. Then the mss is m = M y = x x dx = x 3 dx = 4 so x = = 3 =.5. g-cm = = 8 g x dx = 8 3 while 5. The tringulr region ppers in the mrgin. Here f (x) = h h b x for x b nd g(x) =. The mss is just the re of the tringle, so m = b h while: M y = nd: M x = b b So (x, y) = x h hb ] x dx = b hx hb ] x dx = h x h ] b 3b x3 = b h 6 h h ] ( b x dx = b ) ( h h ) ] 3 b 6 h b x = + b 6h h3 = bh 6 ( ) b h bh ( 6 6 b, = bh bh 3, h ) The centroid of ech region is locted t the point where the lines of symmetry intersect (see mrgin figure).

70 458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss M nd rdius R to height H bove the surfce of the moon: R+H R GMm x dx = GMm x ] R+H R = GMm R GMm R + H Notice tht s the height H grows very lrge, the second term in this nswer becomes very smll nd the totl work pproches GMm R. We cn write: GMm lim H R GMm ] = GMm R + H R Here we re tking limit of n expression tht rose s the vlue of definite integrl, so we cn lso write: GMm GMm = lim R H R GMm ] R+H GMm = lim R + H H R x dx We could write this lst integrl, t lest informlly, s: R GMm x We cll this new type of integrl n improper integrl becuse the intervl of integrtion is infinite, violting n ssumption we mde when dx originlly developing the definite integrl b f (x) dx using Riemnn sums tht the length of the intervl of integrtion,, b], ws finite. Exmple. Represent the re of the infinite region between f (x) = x nd the x-xis for x (see mrgin) s n improper integrl. Solution. We cn represent the re of region (which hs infinite length) s: x dx We don t yet know whether this re is finite or infinite. Prctice. Represent the volume swept out when the infinite region between f (x) = nd the x-xis for x 4 is revolved bout the x-xis x (see mrgin) using n improper integrl. Generl Strtegy for Improper Integrls In the lifting--pylod exmple bove, we defined our first improper integrl s the limit of proper integrl over finite intervl s the length of the intervl becme lrger nd lrger.

71 5.7 improper integrls 459 Our generl pproch to evlute improper integrls over infinitely long intervls s well s nother type of improper integrl introduced lter in this section will mimic this strtegy: Shrink the intervl of integrtion so you hve (proper) definite integrl you cn evlute, then let the intervl grow to pproch the desired intervl of integrtion. The vlue of the improper integrl will be the limiting vlue of the (proper) definite integrls s the intervls grow to the intervl you wnt, provided tht this limit exists. Infinitely Long Intervls of Integrtion To evlute n improper integrl on n infinitely long intervl: replce the infinitely long intervl with finite intervl evlute the integrl on the finite intervl let the finite intervl grow longer nd longer, pproching the originl infinitely long intervl Exmple. Evlute dx (see mrgin). x Solution. The intervl, ) is infinitely long, but we cn evlute the integrl on finite intervls such s, ],, ],, ] nd, more generlly,, M] where M is some mssive positive number: x dx = ] x x dx = x dx = x x nd, more generlly, ] = = ] M ] ] = = = M dx = lim x M ] ] = = 9 ] ] = = 999 x dx = M so: The vlue of the improper integrl is. dx = lim x M ] = M We sy tht the improper integrl dx in the Exmple is x convergent nd tht it converges to. Furthermore, from Exmple, we know tht this improper integrl represents the re of n infinitely long region. We now hve n exmple which you my find highly counterintuitive of region with infinite length but finite re. Not ll improper integrls converge, however.

72 46 pplictions of definite integrls Exmple 3. Evlute ech improper integrl. (See mrgin for grphicl interprettions of these integrls s res of unbounded regions.) () + x dx (b) x dx (c) cos(x) dx Solution. () Replcing the upper limit of the improper integrl with mssive positive number M: M dx = lim + x M = lim M + x dx = lim ] = π rctn(m) M ] M rctn(x) so the improper integrl is convergent nd converges to π. (b) Replcing the upper limit of the improper integrl with mssive positive number M: M ] M dx = lim dx = lim ln(x) x M x = lim ln(m) = M M Becuse this limit diverges, we sy the improper integrl is divergent or tht it diverges. (c) Once gin replcing with M in the upper limit of the integrl: M ] M lim cos(x) dx = lim sin(x) = lim sin(m) M M M As M grows without bound, the vlues of sin(m) oscillte between nd, never pproching single vlue, so the limit does not exist; we sy tht this improper integrl diverges. Prctice. Evlute: () x 3 dx (b) sin(x) dx Definition: For ny integrble function f (x) defined for ll x nd ny integrble function g(x) defined for ll x b: b M f (x) dx = lim M g(x) dx = lim N b N f (x) dx g(x) dx If the limit in question exists nd is finite, we sy tht the corresponding improper integrl converges or is convergent nd define the vlue of the improper integrl to be the vlue of the limit. If the limit in question does not exist, we sy tht the corresponding improper integrl diverges or is divergent.

73 5.7 improper integrls 46 Functions Undefined t n Endpoint of the Intervl of Integrtion Consider the grph of x on the intervl (, ] (see mrgin) nd compre this region to the grph from Exmple. It ppers we cn generte the new region by reflecting the old region cross y = x nd dding rectngle (of re ) t the bottom, so we might resonbly ssume tht the integrl x dx is finite number. This integrl is over finite intervl,, ], but we hve new problem: the integrnd is undefined t x =, one of the endpoints of the intervl of integrtion. This violtes nother ssumption we mde when developing the definition of definite integrl s limit of Riemnn sums. If the function you wnt to integrte is unbounded t one of the endpoints of n intervl of finite length, s in this sitution, you cn shrink the intervl of integrtion so tht the function is bounded t both endpoints of the new, smller intervl, then evlute the integrl over the smller intervl, nd finlly let the smller intervl grow to pproch the originl intervl. Exmple 4. Evlute x dx. Solution. The function x is not defined t x =, the lower endpoint of integrtion, but the function is bounded on intervls such s.36, ],.9, ] nd, more generlly, on the intervl c, ] for ny c > :.36.9 nd, in generl: dx = ] x x dx = x x c ].36 =.36 =. =.8.9 =.9 =.6 =.4 dx = ] x = c = c x c so, tking the limit s c decreses towrd : ] lim dx = lim c = c + c x c + which is wht you should hve expected bsed on the grph. A region of re plus rectngle of re should hve n re of + =. Definition: For ny function f (x) defined nd continuous on (, b] nd ny function g(x) defined nd continuous on, b): b b b f (x) dx = lim c + g(x) dx = lim c b c c f (x) dx g(x) dx

74 46 pplictions of definite integrls See Problems 6 for prctice with integrls of this type. If the limit exists, we sy the integrl converges nd define the vlue of the integrl to be the vlue of the limit. If the limit does not exist, we sy tht the integrl diverges. Prctice 3. Show tht () x dx = 6 nd (b) dx diverges. x If n integrnd is unbounded t one or more points inside the intervl of integrtion, you cn split the originl improper integrl into two or more improper integrls over subintervls where the integrnd is unbounded t only one endpoint of ech subintervl. Testing for Convergence: The P-Test nd the Comprison Test Sometimes we cre only whether or not n improper integrl converges. We now consider two methods for testing the convergence of n improper integrl. Neither method gives you the ctul vlue of the integrl, but ech enbles you to determine whether or not certin improper integrls converge. The Comprison Test for Integrls enbles you to determine the convergence (or divergence) of certin integrls by compring them with other (esier) integrls. The P-Test involves specil cses often used with the Comprison Test for Integrls. P-Test for integrls: For ny >, the improper integrl converges if p > nd diverges if p. x ] = x p dx Proof. It is esiest to consider three cses rther thn two: p =, p > nd p <. If p = then: x p dx = M ] M dx = lim dx = lim ln ( x ) x M M = lim M ln(m) ln() so the improper integrl diverges. For the other two cses, p =, so: M x lim x p p+ ] M M p ] dx = lim = lim M M p + M p p p If p >, then p < so lim M M p p = nd: M p+ xp dx = lim M p + p+ p + ] = p p which is finite number, so the improper integrl converges. If p <, M then p > p so lim M p = nd: xp dx = lim M so the improper integrl diverges. M p+ ] p + p+ = p +

75 5.7 improper integrls 463 Exmple 5. Determine the convergence or divergence of ech integrl. () 5 x dx (b) x dx (c) 8 3 x dx Solution. () The integrl mtches the form required by the P-Test with p = >, so the improper integrl converges. The P-Test does not tell us the vlue of the integrl. (b) The integrl mtches the form required by the P-Test with p = <, so the improper integrl diverges. (c) This is not n improper integrl, so the P-Test does not pply, but: 8 8 ] 3 dx = x dx = x x 3 = 3 ] = 3 4 ] = 9 so the vlue of the integrl is 4.5. The following Comprison Test enbles us to determine the convergence or divergence of n improper integrl of positive function by compring this function with functions whose improper integrls we lredy know converge or diverge. Comprison Test for Integrls of Positive Functions: Suppose f (x) nd g(x) re defined nd integrble for ll x with f (x) g(x). Then: g(x) dx converges f (x) dx converges. f (x) dx diverges g(x) dx diverges. The proof involves strightforwrd ppliction of the definition of n improper integrl nd vrious fcts bout limits, but the grph in the mrgin provides geometriclly intuitive wy of understnding why these results must hold. If g(x) dx converges, then the re under the grph of g(x) is finite, so the (smller) re under the grph of f (x) must lso be finite, nd f (x) dx must converge s well. If f (x) dx diverges, then the re under the grph of f (x) is infinite, so the (bigger) re under the grph of g(x) must lso be infinite, nd g(x) dx must lso diverge. Just s importnt s understnding wht this Comprison Test does tell us is relizing wht the Comprison Test does not tell us. If g(x) dx diverges, or if f (x) dx converges, the Comprison Test tells us bsolutely nothing bout the convergence or divergence of the other integrl. Geometriclly, if g(x) dx diverges, then the re under the grph of g(x) is infinite, but the (smller) re under the

76 464 pplictions of definite integrls grph of f (x) could be either finite or infinite, so we cn t conclude nything bout the convergence or divergence of f (x) dx. Likewise, if f (x) dx converges, then the re under the grph of f (x) is finite, but the (bigger) re under the grph of g(x) could be either finite or infinite, so we cn t conclude nything bout the convergence or divergence of g(x) dx. Exmple 6. Determine whether ech of these integrls is convergent or divergent by compring it with n pproprite integrl tht you lredy know converges or diverges. () 7 x dx (b) Solution. () We know tht 5 > nd x so: 3 + sin(x) 9 x dx (c) dx 6 x 5 x > x 3 < x < x 3 < 7 x < 7 x 3 The numertor of the originl integrnd is constnt nd the dominnt term in the denomintor of tht integrnd is x 3, so it should mke sense to compre the originl integrnd with x 3. The numertor of the originl integrnd fluctutes between nd 4 while the dominnt (nd only) term in its denomintor is x, so it should mke sense to compre the originl integrnd with x. We lso know, by the P-Test with p = 3 >, tht dx con- x3 verges, so 7 x 3 dx = 7 x dx lso converges. By the Com- 3 7 x 3 dx must converge s + 5 prison Test, the smller integrl well. (b) We know tht sin(x), so: 3 + sin(x) 4 < 3 + sin(x) x 4 x = 4 x By the P-Test with p = >, x dx converges, so 4 x dx = 4 dx lso converges. By the Comprison Test, the smller x 3 + sin(x) integrl x dx must converge s well. (c) We know tht u is n incresing function, so: x 5 < x x 5 < x x 5 > x By the P-Test with p = <, integrl 6 6 x dx x 5 dx must lso diverge. diverges, so the bigger

77 5.7 improper integrls Problems In Problems 6, evlute ech improper integrl, or show why it diverges x 3 dx 3 e π π. e 5 x ln(x)] dx + x dx 4. e x dx 5 x ln(x) dx 6. x + x dx x dx 8. (x ) 3 dx. (x + ) dx. x dx 4. 4 x dx x dx 8. sin(x) dx. tn(x) dx. tn(x) dx 4. x + dx 6. π 3 3 (x ) dx x + dx (x + ) 3 dx 3 x dx x dx 3x 8 x 3 dx sin(x) dx x dx x x dx x dx e π x + 5 dx x 3 + x dx 7 x + ln(x) dx + cos(x) x dx 4. x 4 x 5 + dx x + 5 dx x dx x dx x 4 x 6 + dx x 44. x + dx M 45. Exmple 3(b) showed tht dx grew rbitrrily lrge s M grew rbitrrily lrge, so no x finite mount of pint would cover the region bounded by the x-xis nd the grph of f (x) = x for x > : Show tht the volume of the solid obtined when the region grphed bove is revolved bout the x-xis: In Problems 7 44, determine whether ech improper integrl converges or diverges, but do not evlute the integrl x 5 dx 5 x 6 dx x 3 x dx x 7 4 dx x dx 5 x 4 dx x 4 7 dx + x dx is finite, so the 3-dimensionl trumpet-shped region cn be filled with finite mount of pint. Does this present contrdiction?

78 466 pplictions of definite integrls 46. Determine whether or not the volume of the solid obtined by revolving the region between the x- xis nd the grph of f (x) = sin(x) for for x x (see below) bout the x-xis is finite. 5. Use the figure below left to help determine which A A is lrger: x dx or k. k= 47. Compute the volume of the solid obtined when the region in the first qudrnt between the positive x-xis nd the grph of f (x) = x + (see below) is revolved bout the x-xis. 5. Use the figure bove right to help determine A A which is lrger: x dx or k. k= 53. Use the figure below left to help determine which is lrger: A A x dx or k= k. 48. Compute the volume of the solid obtined when the region in the first qudrnt between the positive x-xis nd the grph of f (x) = e x is revolved bout the x-xis. 49. Compute the volume of the solid obtined when the region in the first qudrnt between the positive x-xis nd the grph of f (x) = x + (see below) is revolved bout the y-xis. 5. Compute the volume of the solid obtined when the region in the first qudrnt between the positive x-xis nd the grph of f (x) = e x is revolved bout the y-xis. 54. Use the figure bove right to help determine which is lrger: A x dx or A k= k. The Lplce trnsform of function f (t) is defined using n improper integrl involving prmeter s: F(s) = e st f (t) dt Lplce trnsforms re often used to solve differentil equtions. 55. Compute the Lplce trnsform of the constnt function f (t) =. 56. Compute the Lplce trnsform of f (t) = e 4t. 57. Define function g(t) by: { if t < g(t) = if t Compute the Lplce trnsform of g(t). 58. Define function h(t) by: { if t < 3 h(t) = if t 3 Compute the Lplce trnsform of h(t).

79 5.7 improper integrls 467 b 59. Devise Q-Test to determine whether x q dx converges or diverges for ny number b >. 6. Use the result of the previous problem to test the e π convergence of 3 dx nd x x 3 x dx. 5.7 Prctice Answers ( ). π dx = π 4 x 4 x dx M. () dx = lim x 3 dx = lim ] M x3 M M x = lim M M + ] = (b) Replcing with M in the upper limit of the integrl: sin(x) dx = lim M = lim M M sin(x) dx = lim ] cos(m) + M ] M cos(x) = lim cos(m) M This limit does not exist (the vlues of cos(m) oscillte between nd nd never pproch ny fixed number) so the improper integrl diverges. 3. () The integrl is improper t its upper limit, where x =, so: x dx = c lim ( x) dx = lim c ] = lim c + 9 c c ] c x = + 3 = 6 (b) The integrl is improper t its lower limit, where x =, so: ] dx = lim dx = lim ln ( x ) x c + c x = lim ln() ln(c)] = c + c c + so the integrl diverges.

80 468 pplictions of definite integrls 5.8 Additionl Applictions This section introduces two dditionl pplictions of integrls tht once gin illustrte the process of going from n pplied problem to Riemnn sum nd on to definite integrl. A third ppliction does not follow this process: it uses the ide of re to model n election nd to qulittively understnd why certin election outcomes occur. The min point of this section is to demonstrte the power of definite integrls to solve wide vriety of pplied problems. Ech of these new pplictions is treted more briefly thn those in the previous sections. These re fr from the only pplictions tht could be included here. By now, however, you should hve developed enough of n understnding of the Riemnn-sum process so tht when you encounter other pplictions (in physics, engineering, biology, sttistics, probbility, economics, computer grphics... ) you will be ble to use tht process to set up n integrl to compute or pproximte desired quntity. Fluid Pressures nd Forces In physics, pressure is defined s force per unit of re. The hydrosttic pressure on n object immersed in fluid (such s wter) is the product of the density of tht fluid nd the depth of the object: Fluids exert pressure in ll possible directions, nd the forces due to this pressure ct on solid objects in direction perpendiculr to the object. In the metric system, the stndrd unit of force is pscl (bbrevited P ): P = N m nd nmed fter Blise Pscl (63 66), French mthemticin, physicist, inventor, writer nd philosopher. pressure = (density)(depth) The totl hydrosttic force pplied ginst n immersed object is the sum of the hydrosttic forces ginst ech prt of the object. If n entire object is t the sme depth, we cn determine the totl hydrosttic force ginst tht (necessrily flt) object simply by multiplying the density of the fluid times the depth of the object times the object s re. If the unit of density is pounds per cubic foot nd depth is mesured in feet, then the unit of pressure is pounds per squre foot, mesure of force per unit of re. If pressure, with the units pounds per squre foot, is multiplied by n re with units squre feet, the result is force, mesured in pounds. Exmple. Find the totl hydrosttic force ginst the bottom of the freshwter qurium shown in the mrgin. Solution. Wter s desity is 6.5 lb, so the totl hydrosttic force is: 3 ft ( (density) (depth) (re) = 6.5 lb ) ft 3 (3 ft) ( ft ) or 375 lbs. Finding the totl hydrosttic force ginst the front of the qurium is very different problem, becuse different prts of tht front fce re locted t different depths nd subject to different pressures.

81 5.8 dditionl pplictions 469 To compute the force ginst the front of the qurium, we cn prtition it into n thin horizontl slices (see mrgin) nd focus on one of them. Becuse the slice is very thin, every prt of the k-th slice is t (lmost) the sme depth, so every prt of tht slice is subject to (lmost) the sme pressure. We cn pproximte the totl hydrosttic force ginst the slice t the depth x k s: ( (density) (depth) (re) = 6.5 lb ) ft 3 (x k ft) ( ft) ( x k ft) or 5x k x k lbs. The totl hydrosttic force ginst the front is the sum of the forces ginst ech slice: totl hydrosttic force n 5x k x k k= which is Riemnn sum. The limit of this Riemnn sum s the slices get thinner ( x k ) is definite integrl: n 5x k x k k= x=3 x= 5x dx = 6.5x ] x=3 = 56.5 lbs x= Prctice. Find the totl hydrosttic force ginst one side of the qurium nd the totl force ginst the entire qurium. Exmple. Find the totl hydrosttic force ginst viewing windows A nd B in the freshwter qurium shown in the mrgin. Solution. For window A, using similr tringles, the width w of slice t depth x m stisfies: w 6 x = 3 w = 3 (6 x) = 9 3 x so the re of slice of height x k m t depth x k m is ( 9 3 x k) xk m. The density of wter is kg m. Multiplying this density by re (with 3 units m ) would give kg per m, but pressure is mesured in N per m, so we need to multiply by the ccelertion due to grvity, g 9.8 sec m. The hydrosttic force pplied to the k-th slice is thus: (9.8)x k (9 3 ) x k x k nd the totl hydrosttic force pplied to the window is therefore: x=6 x=4 98 9x 3 x ] dx = 98 9 x x3 ] 6 4 = 98 (6 8) (7 3)] = 3734 N

82 47 pplictions of definite integrls For window B, pplying the Pythgoren Theorem yields: ( w ) (5 x) + = w = (5 x) The totl hydrosttic force is thus: x=6 x=4 (9.8)x (5 x) dx which (using technology) is pproximtely 54,95 N. Prctice. Find the totl hydrosttic force ginst viewing windows C nd D of the freshwter qurium shown in the mrgin. Becuse the totl force t even moderte depths is so lrge, underwter windows re mde of thick glss or plstic nd strongly secured to their frmes. Similrly, the bottom of dm is much thicker thn the top in order to withstnd the greter force ginst the bottom. Kinetic Energy Physicists define the kinetic energy (energy of motion) of n object with mss m nd velocity v to be: KE = m v The greter the mss of n object or the fster it is moves, the greter its kinetic energy. If every prt of the object hs the sme velocity, computing its kinetic energy becomes reltively esy. Sometimes, however, different prts of n object move with different velocities. For exmple, if n ice skter is spinning with n ngulr velocity of revolutions per second, her rms trvel further in one second (hve greter liner velocity) when they re extended thn when drwn in close to her body (see mrgin). So the ice skter, spinning t revolutions per second, hs greter kinetic energy when her rms re extended. Similrly, the tip of rotting propeller (or the brrel of swinging bsebll bt) hs greter liner velocity thn other prts of the propeller (or the bt s hndle). If the units of mss re kg nd the units of velocity re m/sec, then: KE = ( (m kg) v m ) m = sec mv kg m sec so the units of kinetic energy re N-m, or Joules, the sme s work. Similrly, if the units of mss re g nd the units of velocity re cm/sec, then the units of kinetic energy re dyn-cm, or ergs. Exmple 3. A point-mss of grm t the end of (mssless) -cm string rottes t rte of revolutions per second (see mrgin).

83 5.8 dditionl pplictions 47 () Find the kinetic energy of the point-mss. (b) Find its kinetic energy if the string is cm long. Solution. () In one second, the mss trvels twice round circle with rdius cm so it trvels (π ) = 4π cm. Its velocity is thus v = 4π cm/sec, nd: KE = mv = ( ( g) 4π cm ) = 8π ergs sec or bout.79 J. (b) If the string is cm long, then the velocity is (π 4) = 8π cm/sec nd: KE = mv = ( ( g) 8π cm ) = 3π ergs sec or bout.36 J. When the length of the string doubles, the velocity doubles nd the kinetic energy qudruples. Prctice 3. A -grm point-mss t the end of -meter (mssless) string rottes t rte of 4 revolutions per second. Find the kinetic energy of the point mss. If different prts of rotting object re different distnces from the xis of rottion, then those prts hve different liner velocities, nd it becomes more difficult to clculte the totl kinetic energy of the object. By now the method should seem very fmilir: prtition the object into smll pieces, pproximte the kinetic energy of ech piece, nd dd the kinetic energies of the smll pieces ( Riemnn sum) to pproximte the totl kinetic energy of the object. The limit of the Riemnn sum s the pieces get smller is definite integrl. Exmple 4. The density of nrrow br (see mrgin) is 5 grms per meter of length. Find the kinetic energy of the 3-meter-long br when it rottes t rte of revolutions per second. Solution. Prtition the br (see mrgin) into n pieces so tht the mss of the k-th piece is: ( m k (length) (density) = ( x k m) 5 g ) = 5 x m k g During one second, the k-th piece, locted t distnce of x k m from the pivot line, will mke two revolutions, trveling pproximtely: ] (π rdius]) = 4π x k cm = 4πx k cm so v k 4πx k cm/sec. The kinetic energy of the k-th piece is: m k v k ( (5 x cm ) k g) 4πx k = 4π xk sec ergs

84 47 pplictions of definite integrls nd the totl kinetic energy of the rotting br is therefore: n 4π xk x k k= x=3 x= 4π x dx = 4π 3 x3 ] 3 which equls 36π ergs, or bout 3.55 J. Prctice 4. Find the kinetic energy of the br in the previous Exmple if it rottes t revolutions per second t the end of -centimeter (mssless) string (see mrgin). Exmple 5. Find the kinetic energy of the thin, flt object with density.7 g/cm shown in the mrgin when it rottes t 45 revolutions per minute. Solution. We cn prtition the object long one rdil line nd form n nnulr slices ech x cm wide. Then the slice between x k nd x k + x is thin nnulus ( disk with smller disk removed from its center) with re: ] π (x k + x) π (x k ) = π xk + x k x + ( x) xk = πx k x + π ( x) πx k x The slices tht give rise to the Riemnn sum in this problem re unlike most exmples we hve seen previously not rectngles. We lso use here the notion tht if x is smll, then ( x) is very smll, so we cn essentilly ignore it in our pproximtion of re. nd mss (.7)πx k x. During one revolution, point on this slice trvels pproximtely πx k cm nd 45 rev/min is equivlent to 3 4 rev/sec, so the liner velocity of the point is πx k 34 = 3 πx k cm/sec. The kinetic energy of this slice is therefore: ( ) ( ) 3 (.7)πx k x πx k = 9 (.7)π xk 3 x so the totl kinetic energy of the object is: n k= 9 (.7)π x 3 k x b Evluting this integrl yields: 9 (.7)π x 3 dx ] b 9 (.7)π 4 x4 = 9 8 (.7)π b 4 4] In the not-so-distnt pst your grndprents (nd perhps even your prents) used such objects to listen to music nd ech one only held two songs! Becuse b is rised to the fourth power, smll increse in the vlue of b (if b > ) leds to lrge increse in the object s kinetic energy. If =.75 in.95 cm nd b = 3.75 in 9.55 cm, the totl mss of the object is 4 g nd its totl kinetic energy is bout 5,5 ergs.

85 5.8 dditionl pplictions 473 Ares nd Elections The previous pplictions in this chpter hve used definite integrls to determine res, volumes, pressures nd energies precisely. But exctness nd numericl precision re not the sme s understnding, nd sometimes we cn gin insight nd understnding simply by determining which of two res or integrls is lrger. One sitution of this type involves models of elections. Suppose the voters of stte hve been surveyed bout their positions on single issue, with their responses recorded on quntittive scle. The distribution of voters who plce themselves t ech position on this issue ppers in the mrgin. Suppose lso tht ech voter csts his or her vote for the cndidte whose position on this issue is closest to his or her position. If two cndidtes hve tken the positions lbeled A nd B, then voter t position c votes for the cndidte t A becuse A is closer to c thn B is to c. Similrly, voter t position d votes for the cndidte t B. The totl votes for the cndidte t A in this election is represented by the shded re under the curve, nd the cndidte with the lrger number of votes the lrger re wins the election. In this illustrtion, the cndidte t A wins. Exmple 6. The distribution of voters on n issue ppers below left. If these voters decide between cndidtes on the bsis of tht single issue, which cndidte will win the election? Solution. The figure bove right illustrtes tht A corresponds to lrger re (more votes) thn B: A will win. Prctice 5. In n election between cndidtes with positions A nd B in the mrgin figure, who will win? If voters behve s described nd if the election is between two cndidtes, then we cn give the cndidtes some dvice. The best position for cndidte is t the medin point, the loction tht divides the voters into two equl-sized (equl-re) groups so tht hlf of the voters re on one side of the medin point nd hlf re on the other side (see mrgin). A cndidte t the medin point gets more votes thn cndidte t ny other point. (Why?) If two cndidtes hve positions on opposite sides of the medin point (see mrgin), then cndidte cn get more votes by moving bit towrd the medin point. This move towrd the middle ground

86 474 pplictions of definite integrls commonly occurs in elections s cndidtes ttempt to sell themselves s modertes nd their opponents s extremists. If more thn two cndidtes re running in n election, the sitution chnges drmticlly. A cndidte t the medin position, the unbetble plce in two-cndidte election cn even get the fewest votes. If the mrgin figure represents the distribution of voters on the single issue in the election, then cndidte A would bet B in n election just between A nd B (below left) nd A would bet C in n election just between A nd C (below center). But in n election mong ll three cndidtes, A would get the fewest votes (below right). This type of sitution relly does occur. It leds to the politicl sying bout primry election with mny cndidtes nd generl election between the finl nominees from two prties: extremists cn win primries, but modertes re elected to office. The previous discussion of elections nd res is gretly oversimplified. Most elections involve severl issues of different importnce to different voters, nd the views of the voters re seldom completely known before the election. Mny cndidtes tke fuzzy positions on issues. And it is not even certin tht rel voters vote for the cndidte with the closest position: perhps they don t vote t ll unless some cndidte is close enough to their position. But this very simple model of elections cn still help us understnd how nd why some things hppen in elections. It is lso strting plce for building more sophisticted models to help understnd more complicted election situtions nd to test ssumptions bout how voters relly do mke voting decisions.

87 5.8 dditionl pplictions Problems In Problems 5, use ρ for the density of the fluid in the given continer.. Clculte the force ginst windows A nd B in the figure below. 5. Clculte the totl force ginst the end of the tnk shown below. 6. The three tnks shown below re ll 6 feet tll nd the top perimeter of ech tnk is feet. Which tnk hs the gretest totl force ginst its sides?. Clculte the force ginst windows C nd D in the figure bove. 3. Clculte the totl force ginst ech end of the tnk shown below. How does the totl force ginst the ends of the tnk chnge if the length of the tnk is doubled? 7. The three tnks shown below re ll 6 feet tll nd the cross-sectionl re of ech tnk is 6 ft. Which tnk hs the gretest totl force ginst its sides? 4. Clculte the totl force ginst ech end of the tnk shown below. 8. Clculte the totl force ginst the bottom feet of the sides of tnk with squre 4-foot by 4- foot bse tht is filled with wter () to depth of 3 feet. (b) to depth of 35 feet. 9. Clculte the totl force ginst the bottom feet of the side of cylindricl tnk with rdius of feet tht is filled with wter () to depth of 3 feet. (b) to depth of 35 feet.

88 476 pplictions of definite integrls. Clculte the totl force ginst the side nd bottom of cylindricl luminum sod cn with dimeter 6 cm nd height cm if it is filled with 385 g of sod. (Assume the cn hs been opened so crboniztion is not fctor.). Find the kinetic energy of -grm object rotting t 3 revolutions per second t the end of () 5-cm (mssless) string nd (b) -cm string.. Ech centimeter of metl br hs mss of 3 grms. Clculte the kinetic energy of the 5- centimeter br if it is rotting t rte of revolutions per second bout one of its ends. 3. Ech centimeter of metl br hs mss of 3 grms. Clculte the kinetic energy of the 5- centimeter br if it is rotting t rte of revolutions per second t the end of -cm cble. 4. Clculte the kinetic energy of -grm meter stick if it is rotting t rte of revolution per second bout one of its ends. 5. Clculte the kinetic energy of -grm meter stick if it is rotting t rte of revolution per second bout its center point. 6. A flt, circulr plte is mde from mteril tht hs density of grms per cubic centimeter. The plte is 5 centimeters thick, hs rdius of 3 centimeters nd is rotting bout its center t rte of revolutions per second. () Clculte its kinetic energy. (b) Find the rdius of plte tht would hve twice the kinetic energy of the first plte, ssuming the density, thickness nd rottion rte re the sme. 7. Ech wsher in the figure below is mde from mteril with density of grm per cm 3, nd ech is rotting bout its center t rte of 3 revolutions per second. Clculte the kinetic energy of ech wsher (dimensions re in cm). 8. The rectngulr plte shown below is cm thick, cm long nd 6 cm wide nd is mde of mteril with density of 3 grms per cm 3. Clculte the kinetic energy of of the plte if it is rotting t rte of revolutions per second () bout its -cm side nd (b) bout its 6-cm side. 9. Clculte the kinetic energy of the plte in Problem 8 if it is rotting t rte of revolutions per second bout verticl line through the center of the plte, s shown below left.. Clculte the kinetic energy of the plte in Problem 8 if it is rotting t rte of revolutions per second bout verticl line through the center of the plte, s shown bove right.. For the voter distribution shown below, which cndidtes would the voters t positions, b nd c vote for?

89 5.8 dditionl pplictions 477. For the voter distribution shown below, which cndidtes would the voters t positions, b nd c vote for? 6. Refer to the voter distribution shown below. 3. Shde the region representing votes for cndidte A in the distribution shown below. Which cndidte wins? () Which cndidte wins? (b) If cndidte B withdrws before the election, which cndidte will win? (c) If cndidte B stys in the election but C withdrws, then who wins? 7. Refer to the voter distribution shown below. 4. Shde the region representing votes for cndidte A in the distribution shown below. Which cndidte wins? () If the election is between A nd B, who wins? (b) If the election is between A nd C, who wins? (c) If the election is mong A, B nd C, who wins? 8. Refer to the voter distribution shown below. 5. Refer to the voter distribution shown below. () Which cndidte wins? (b) If cndidte B withdrws before the election, which cndidte will win? (c) If cndidte B stys in the election but C withdrws, then who wins? () If the election is between A nd B, who wins? (b) If the election is between A nd C, who wins? (c) If the election is mong A, B nd C, who wins? 9. Sketch distribution for two-issue election.

90 478 pplictions of definite integrls 5.8 Prctice Answers. The resoning for side of the qurium is exctly the sme s for the front, except side is foot long insted of, so the force is hlf of tht ginst the front: 8.5 lbs. The totl force ginst ll sides (nd the bottom) is: (8.5) + (56.5) = 6.5 lbs. For window C, using similr tringles (see mrgin), the width w of slice t depth x m stisfies: w x 4 = 3 w = 3 (x 4) = 3 x 6 so the re of slice of height x k m t depth x k m is ( 3 x k 6 ) x k m. The density of wter is kg m nd g 9.8 m 3 sec, so the hydrosttic force pplied to the k-th slice is: ( ) 3 (9.8)x k x k 6 x k nd the totl hydrosttic force pplied to the window is therefore: x=6 x=4 ] ] x 6x dx = 98 x3 3x 4 ] = 98 (8 8) (3 48) = 5696 N Windows A (from Exmple ) nd C (with horizontl flip) fit together to form window D, so it is encourging tht the sum of the totl hydrosttic forces ginst A nd C is = 943 N, the totl hydrosttic force ginst window D. For window D, the width is 3 t ll depths, so the totl hydrosttic force ginst the window is: x=6 x=4 6 (9.8)x 3 dx = 475x = 943 N 4 3. The object trvels π ( m) = 4π m during one revolution, so during the second it tkes to mke 4 revolutions, the object trvels 6π m; its velocity is thus v = 6π sec cm nd its kinetic energy is: m v = ( ( g) 6π cm ) = 8π ergs ergs sec 4. Everything remins the sme s in Exmple 4, except for the endpoints of integrtion: x=4 x= 4π x dx = 4π 3 x3 ] 4 = 84π ergs which is pproximtely 8, 94, 677 ergs, or 8.9 J. 5. The shded regions in the mrgin figure show the totl votes for ech cndidte: B wins.

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