5.2 Volumes: Disks and Washers

Transcription

1 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 y-plne bout horizontl or verticl line. We cll solid formed in this wy solid of revolution nd we cll the line n is of rottion. If the is of rottion coincides with boundry of the region (s in the mrgin figure) then the cross-sections of the region perpendiculr to the is of rottion will be disks, mking it reltively esy to find formul for the re of cross-section: A() = re of disk (rdius) The rdius is often function of, the loction of the cross-section. Emple. Find the volume of the solid (shown in the mrgin) formed by rotting the region in the first qudrnt bounded by the curve y = nd the line = 4 bout the -is. Solution. Any slice perpendiculr to the -is (nd to the y-plne) will yield circulr cross-section with rdius equl to the distnce between the curve y = nd the -is, so the volume of the region is given by: π d = π 4 d 8 4 = π or bout 6.8 cubic inches. Sometimes the boundry curve intersects the is of rottion. Emple. The region between the grph of f () = nd the horizontl line y = for 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, ll of them disks. If, then the rdius of the disk is r() = ; if, then r() =. We could split up the volume computtion into two seprte integrls, using A() r() = π for nd A() r() for, but: π ( )

2 5. volumes: disks nd wshers 4 for ll so we cn insted compute the volume with single integrl: π d 4 + d = 46π 5 or bout Prctice. Find the volume of the solid formed by revolving the region between f () = 3 nd the horizontl line y = bout the line y = for 3 (see mrgin). Volumes of Revolved Regions ( Disk Method ) If then the region constrined by the grph of y = f (), 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: A() d = π (rdius) d = π f () L d 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 pproimte the re of the region) results in disk. If the region between the grph of f nd the -is (L = ) is revolved bout the -is, then the previous formul reduces to: π f () d Emple 3. Find the volume generted when the region between one rch of the sine curve (for π) nd () the -is is revolved bout the -is 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(): π sin() d 4 sin() π sin () d cos() d π π 4.93

3 4 pplictions of definite integrls (b) Here we use the more generl disk formul with L = : V sin() d sin () sin() + d 4 cos() sin() + d π sin() + cos() 4 3π 4 π + = 3π 4 π or pproimtely.. Prctice. Find the volume generted when () the region between the prbol y = (for ) nd the -is is revolved bout the -is nd (b) the region between the prbol y = (for ) nd the line y = is revolved bout the line y =. Emple 4. Given tht f () d = 4 nd f () d = 7, represent the volume of ech solid shown in the mrgin s definite integrl, nd evlute those integrls. Solution. () Here the is of rottion is y = so: π (rdius) d = (b) Here the is of rottion is y = so: π (rdius) d = π f () d f () d = 7π f () + d π f () ( ) d ( f ()) d + f () d (5 ) = 9π ( f ()) + f () + d d (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: The lst one is left for you. f () π d f () d = 7π 4 Prctice 3. Set up nd evlute n integrl to compute the volume of the lst solid shown in the mrgin.

4 5. volumes: disks nd wshers 43 Solids with Holes Some solids hve holes : for emple, 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 net 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 is tht does not intersect the region. Emple 5. Compute the volume of the solid shown in the mrgin. Solution. The fce for slice mde t hs re: A() = re of BIG circle re of smll circle BIG rdius π smll rdius Here the BIG rdius is the distnce from the line y = + to the -is, or R() = ( + ) = + ; similrly, the smll rdius is the distnce from the curve y = to the -is, or r() = =, hence the cross-sectionl re is: A() + π + + The curves intersect where: = + = + = = ± 4( ) = ± 5 Clerly we need > for this region, so the left endpoint of integrtion must be = + 5 while the right endpoint is =, so the volume of the solid is: V + + d which simplifies to π ( π ) 3 ( )

5 44 pplictions of definite integrls The previous Emple etends 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 () nd y = g() nd the intervl, b is revolved bout horizontl line the volume of the resulting solid is: π (R()) π (r()) d where R() represents the distnce from the is of rottion to the frthest curve from tht is, nd r() represents the distnce from the is to the closest curve. If r() =, 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 is 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 pproimte the volume of the solid. You might lso need to use numericl methods to locte where the boundry curves of the region intersect. Emple 6. Find the volume of the solid generted by rotting the region between the curves y = nd y = bout the () -is (b) y-is (c) the line = (d) the line y = 5. Solution. () The curves intersect where = = ( ) =, so the limits of integrtion should involve = nd =. Revolving verticl slice of the region with width bout the -is yields wsher with big rdius R() = = (the line y = is frthest from the -is) nd smll rdius r() =

6 5. volumes: disks nd wshers 45 = (the prbol is closest to the -is when ). So the volume of the solid is: V () π( ) d 4 4 d π = 64π 5 (b) A verticl slice revolved round the y-is does not result in wsher so insted we try slicing horizontlly. A horizontl slice of thickness y revolved round the y-is does result in wsher. The big rdius is the -distnce from the prbol (where = y) to the y-is (where = ) so R(y) = y. Similrly, the smll rdius is the distnce from the line (where = y ) to the y-is (where = ), 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, = y = ; t the upper intersection point, = y = = = 4. So the volume of the solid is: y=4 V ( ( y) y ) π dy y 4 y dy y= y y = 8π 3 3 (c) This solid resembles the one from prt (b), ecept now the rdii re both bigger becuse the region (nd the curves tht form the boundry of the region) re frther wy from the is of rottion: R() = y ( ) = y + nd r() = y ( ) = y + : y=4 V ( ( y + ) y ) π + dy y= (y + ( y + ) y 4 y dy ) 4 y + y + dy 4 3 y 3 4 y = 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() = 5 nd r() = 5 : = V (5 ) (5 ) d = 36π 5 = 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 = nd y = bout the () the line = 5 (b) the line y = 5.

7 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 -is.. y =, = 5. y = sin(), 3. y = cos(), 3 4. y = 3 5. y = 7 6. y = y = 5 8. = y 9 3. Use clculus to determine the volume of sphere of rdius r. (Revolve the region bounded by the -is nd the top hlf of the circle + y = r bout the -is.) 33. Compute the volume swept out when the top hlf of the ellipticl region bounded by 5 + y 3 = is revolved round the -is (see figure below). 9. = y. + y = y = y = 5 In Problems 3 3, compute the volume of the solid formed when the region between the given curves is rotted bout the specified is. 3. y =, y = 4 bout the -is 4. y =, y = 4 bout the y-is 5. y =, y = 4 bout the y-is 6. y =, y = 4 bout the -is 7. y =, y = 3 bout the -is 8. y = sec(), y = cos(), 3 bout the -is 9. y = sec(), y = cos(), 3 bout the -is. y =, y = 4 bout y = 3. y =, y = 4 bout y = 4. y =, y = 4 bout = 4 3. y =, y = 4 bout = 3 4. y =, y = 4 bout = 5. y = sin(), y =, = bout y = 3 6. y = sin(), y =, bout y = 7. y =, y = 3, bout = 8. y =, y = 3, bout = 4 9. y =, y = 3, bout y = 3. y =, y = 3, bout y = 3 3. Use clculus to compute the volume of sphere of rdius. (A sphere is formed when the region bounded by the -is nd the top hlf of the circle + y = is revolved bout the -is.) 34. Compute the volume swept out when the top hlf of the ellipticl region bounded by + y b = is revolved round the -is. 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.

8 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 -is bout the -is. 39. () Find the re between f () = nd the -is for, nd M. Wht is the limit of the re for M when M? (b) Find the volume swept out when the region in prt () is revolved bout the -is for, nd M. Wht is the limit of the volume for M when M? 5. Prctice Answers. 3 3 π (3 ) ) d ( ) d. () Slicing the region verticlly nd rotting the slice bout the -is results in disks, so the volume of the solid is: 3. π d 3 4 d 5 5 = 3π 5 (b) Here the slices etend from y = to y = so the rdius of ech disk is nd the volume is: π d + d = 3π d = 56π 5 π 3 f () d 9 6 f () + ( f ()) d 5 9 d 6 f () d + f () d = 9π 4. The volume we wnt cn be obtined by subtrcting the volume of the bo from the volume of the truncted cone generted by the rotted line segment. The volume of the truncted cone is: π + d d = 56π 3 while the volume of the bo is () = 4 so the volume of the solid shown in the grph is 56π

9 48 pplictions of definite integrls 5. () Slicing the region verticlly nd rotting the slice bout the line = 5 results in something other thn wsher, so we insted slice the region horizontlly. The slice etends from = y (frthest from the is of rottion) to = y (closest), so the volume of the solid is: ( 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 ) π (5 ) d = 36π 5