Chapter 7: Applications of Integrals


 Clifford Shaw
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1 Chpter 7: Applictions of Integrls 78
2 Chpter 7 Overview: Applictions of Integrls Clculus, like most mthemticl fields, egn with tring to solve everd prolems. The theor nd opertions were formlized lter. As erl s 70 c, Archimedes ws working on the prolem of prolem of finding the volume of nonregulr shpes. Beond his thtu incident tht reveled the reltionship etween weight volume nd displcement, He hd ctull egun to formlize the limiting process to eplore the volume of digonl slice of clinder. This is where Clculus cn give us some ver powerful tools. In geometr, we cn find lengths of specific ojects like line segments or rcs of circles, while in Clculus we cn find the length of n rc tht we cn represent with n eqution. The sme is true with re nd volume. Geometr is ver limited, while Clculus is ver openended in the prolems it cn solve. On the net pge is n illustrtion of the difference. In this chpter, we will investigte wht hve ecome the stndrd pplictions of the integrl: Are Volumes of Rottion Volume CrossSection Arc Length Just s AP, we will emphsize how the formuls relte to the geometr of the prolems nd the technologicl (grphing clcultor) solutions rther thn the lgeric solutions. Below is n illustrtion of wht we cn ccomplish with Clculus s opposed to geometr: 79
3 Geometr Clculus Length Are Volume Clculus cn lso e used to generte surfce res for oddshped solids s well, ut tht is out of the scope of this clss. 80
4 7. Are Between Two Curves We hve lerned tht the re under curve is ssocited with n integrl. But wht out the re etween two curves? It turns out tht it is simple proposition, ver similr to some prolems we encountered in geometr. In geometr, if we wnted to know the re of shded region tht ws composed of multiple figures, like the illustrtion elow, we find the re of the lrger nd sutrct the re of the smller. The re for this figure would e s ATotl Asqure Acircle s π = = Similrl, since the integrl gets us numeric vlue for the re etween curve nd n is, if we simpl sutrct the smller curve from the lrger one nd integrte, we cn find the re etween two curves. Ojectives: Find the re of the region etween two curves. 8
5 Unlike efore, when we hd to e concerned out positives nd negtives from definite integrl messing up our interprettion of re, the sutrction tkes cre of the negtive vlues for us (if one curve is under the is, the sutrction mkes the negtive vlue of the integrl into the positive vlue of the re. Are Between Two Curves: The re of the region ounded the curves = f ( ) nd g( ) = nd the lines = nd = where f nd g re continuous nd f g for ll in, is ( ) ( ) A = f g d You cn lso think of this epression s the top curve minus the ottom curve. If we ssocite integrls s re under curve we re finding the re under the top curve, sutrcting the re under the ottom curve, nd tht leves us with the re etween the two curves. The re of the region ounded the curves = f ( ), g( ) =, nd the lines = c nd = d where f nd g re continuous nd f g for ll in cd, is d c ( ) ( ) A = f g d You cn lso think of this epression s the right curve minus the left curve. Steps to Finding the Are of Region:. Drw picture of the region.. Sketch Reimnn rectngle if the rectngle is verticl our integrl will hve d in it, if our rectngle is horizontl our integrl will hve d in it.. Determine n epression representing the length of the rectngle: (top ottom) or (right left).. Determine the endpoints the oundries (points of intersection). 5. Set up n integrl contining the limits of integrtion, which re the numers found in Step, nd the integrnd, which is the epression found in Step. 6. Solve the integrl. 8
6 E Find the re of the region ounded = sin, = cos, π =, nd = π. 5 6 ( ) ( ) A = f g d Our pieces re perpendiculr to the is, so our integrnd will contin d sin ( ) cos( ) On our intervl sin ( ) is greter thn cos( ) π π sin ( ) cos( ) = to = π A= d A = d Our region etends from π A= cos ( ) sin( ) A= ( ( π) ( π) ) π π π π cos sin cos sin = 8
7 Wht if the functions re not defined in terms of ut in terms of insted? E Find the re of the region ounded = e, =, =. =, nd d ( ) ( ) A = f c g d Our pieces re perpendiculr to the is, so our integrnd will contin d ( ) ( ) d A= e d c = On our intervl e is greter thn A e d Our region etends from A=.959 = to = 8
8 π E Find the re of the region ounded = sin, = cos, = 0, nd = π π A ( ) ( ) d ( ) ( ) d 0 π = cos sin + sin cos =.88 With this prolem, we hd to split it into two integrls, ecuse the top nd ottom curves switch prtw through the region. We could hve lso done this with n solute vlue: π A ( ) ( ) d 0 = cos sin =.88 If we re sked this question on n AP test or in college clssroom, the usull wnt to see the setup. The solute vlue is just n es w to integrte using the clcultor. NOTE: An integrl cn e solved going d or d. It is usull the cse though, tht one is esier thn the other. 85
9 E Sketch the grphs of = 6 nd =. If sked to find the re of the region ounded the two curves which integrl would ou choose n integrl with dnd n integrl with d? Both will work ut one is less time consuming We cn see tht if we went d we would need three integrls (ecuse the tops nd ottoms re different on three different sections), ut if we go d we onl need one integrl. I like integrls s much s the net person, ut I ll just do the one integrl if tht s OK right now. d c ( ) ( ) A = f g d To find the oundries, we need to set the equtions equl to ech other: = 6 0 = 0 0 = 5 = 0,5 ( ) 86
10 5 0 5 ( ) ( 6 ) A= d = ( 0 )d 0 = 5 = A = 87
11 7. Homework Drw nd find the re of the region enclosed the given curves.. = +, = 9, =, = π. = sin, = e, = 0, = =, = 8, =, =.. π = sin, = cos, = 0, = 5. = 5, = 6. + = +, = 7. = e, =, =, = =, = = +, =, =, = 0. =, = Use our grpher to sketch the regions descried elow. Find the points of intersection nd find the re of the region descried region.. =, = e, =. ( ) = ln +, = cos. =, = 88
12 7. Volume Rottion out n Ais We know how to find the volume of mn ojects (rememer those geometr formuls?) i.e. cues, spheres, cones, clinders ut wht out nonregulr shped oject? How do we find the volumes of these solids? Luckil we hve Clculus, ut the sis of the Clculus formul is ctull geometr. If we tke function (like the prol elow) nd rotte it round the is on n intervl, we get shpe tht does not look like nthing we could solve with geometr. When we rotte this curve out the is, we get shpe like the one elow: z 89
13 This is sort of like cone, ut ecuse the surfce curves inwrd its volume would e less thn cone with the sme size se. In fct, we cnnot use simple geometr formul to find the volume. But if we egin thinking of the curve s hving Reimnn rectngles, nd rotte those rectngles, the prolem ecomes little more ovious: Ech of the rotted rectngles ecomes clinder. The volume of clinder is V = π r h Note tht the rdius (r) is the distnce from the is to the curve (which is f ( )) nd the height is the chnge in ( Δ ), giving us volume of ( ( )) V = π f Δ 90
14 To find the totl volume, we would simpl dd up ll of the individul volumes on the intervl t which we re looking (let s s from = to = ). totl = ( ( )) V = π f Δ Of course, this would onl give us n pproimtion of the volume. But if we mde the rectngles ver nrrow, the height of our clinders chnges from Δ to d, nd to dd up this infinite numer of terms with these infinitel thin clinders, the ecomes = giving us the formul totl π ( ) ( ) V = f d Wht ou m relize is tht the f ( ) is simpl the rdius of our clinder, nd we usull write the formul with n r insted. The reson for this will ecome more ovious when we rotte out n is tht is not n  or is. And since integrtion works whether ou hve d or d, this sme process works for curves tht re rotted round the is nd re defined in terms of s well. Volume Disc Method (Prt ): The volume of the solid generted when the f, is rotted out the is function f ( ) from = nd =, where ( ) 0 [or g( ) from = c nd = d, where ( ) 0 given g, is rotted out the is is or = π ( ) or π ( ) V f d V = π d c V = g d r d where r is the height (or length if the re horizontl) of our Riemnn rectngle. 9
15 Steps to Finding the Volume of Solid: Ojective:. Drw picture of the region to e rotted.. Drw the is of rottion.. Sketch Reimnn rectngle if the piece is verticl our integrl will hve d in it, if our piece is horizontl our integrl will hve d in it. Your rectngle should lws e sketched perpendiculr to the is of rottion.. Determine n epression representing the rdius of the rectngle (in these cses, the rdius is the function vlue). 5. Determine the endpoints the region covers. 6. Set up n integrl contining π outside the integrnd, the limits of integrtion re the numers found in Step 5, nd the integrnd is the epression found in Step. Find the volume of solid rotted when region is rotted out given line. 9
16 E Let R e the region ounded the equtions =, the is, =, nd = 7. Find the volume of the solid generted when R is rotted out the is V r d = π Our pieces re perpendiculr to the is, so our integrnd will contin d V = π d We cnnot integrte with respect to so we will V π d V 7 π d sustitute out for = The epression for is = Our region etends from = to = 7 V = 55π When the region, R, is rotted s in the emple ove, the solid generted would look like this: 9
17 z Note tht knowing tht this is wht the rotted solid looks like hs no ering on the mth. It mkes n interesting shpe, ut eing le to drw the imge nd eing le to generte the volume integrtion re two completel seprte things. We re much more concerned with finding the volume. 9
18 π E Let R e the region ounded the curves = sec, the is, =, 5π nd =. Find the volume of the solid generted when R is rotted out the is. π/ π π/ π 5π/ V V r d = π Our pieces re perpendiculr to the is, so our d integrnd will contin d = π We cnnot integrte with respect to so we will sustitute out for V = π sec d The epression for is sec 5π π V = π sec d Our region etends from V = π π 5π = to = 95
19 Agin, when the region, S, is rotted s in the emple ove, the solid generted would look like this: z It s prett coollooking, ut of no gret consequence in terms of the mth. 96
20 Wht if we hd function defined in terms of? E Find the volume of the solid otined rotting out the is the region ounded = nd the is. V V d r d c = π Our pieces re perpendiculr to the is, so our d d c integrnd will contin d = π We cnnot integrte with respect to so we will V d = π d c V = π d 0 su out for The epression for is Our region etends from = 0 to = 6π V = 5 97
21 Suppose we wnted to find the volume of region ounded two curves tht ws then rotted out n is. Volume Wsher Method: The volume of the solid generted when the region f g [or ounded functions f ( ) nd g( ), from = nd =, where ( ) ( ) f ( ) nd g( ), from = c nd = d, where f ( ) g( ) is is given π ], is rotted out the ( ) ( ) or = ( ) ( ) V = f g d or d π c V f g d d ( ) or = π ( ) V = π R r d c V R r d Where R is the outer rdius of our Riemnn rectngle nd r is the inner rdius of our Riemnn rectngle. 98
22 Agin, think of tking tin strip of this region, Riemnn rectngle, nd rotting it out the is. Note tht this gives us clinder (leit ver nrrow one) with nother clinder cut out of it. This is wh the formul is wht we stted ove. If we took ll of those strips for the ove emple, the solid would look like this. z Notice tht overll, this does not look like wsher per se, ut if ou cut cross section, ou would see wsher shpes like the one illustrted elow: Cross section of the solid ove: 99
23 Steps to Finding the Volume of Solid With the Wsher Method:. Drw picture of the region to e rotted.. Drw the is of rottion.. Sketch Riemnn rectngle if the piece is verticl our integrl will hve d in it, if our piece is horizontl our integrl will hve d in it. Your rectngle should lws e sketched perpendiculr to the is of rottion.. Determine n epression representing the rdius of the rectngle in the cse of wshers ou will hve n epression for the outer rdius nd n epression for the inner rdius. 5. Determine the oundries the region covers. 6. Set up n integrl contining π outside the integrnd, the limits of integrtion re the oundries found in Step 5, nd the integrnd is the epression found in Step. E 5 Let R e the region ounded the equtions = nd =. Find the volume of the solid generted when R is rotted out the is R r
24 ( ) V = π R r d Our pieces re perpendiculr to the is, so our integrnd will contin d π ( ) ( ) V = d The epression for R is nd the epression π ( ) ( ) 0 for R is V = d Our region etends from = 0 to = V =.9 E 6 Let R e the region ounded =, =, nd = 0. Find the volume of the solid generted when R is rotted out the is R r ( ) V = π R r d Our pieces re perpendiculr to the is, so our integrnd will contin d π ( ) ( ) V = d The epression for R is nd the epression π ( ) ( ) 0 for R is V = d Our region etends from = 0 to = V =.57 0
25 E 7 Let R e the region ounded =, =, nd = 0. Find the volume of the solid generted when R is rotted out the is R r d ( ) V = π c R r d Our pieces re perpendiculr to the is, so our integrnd will contin d d c ( ) ( ) V = π d The epression for R is nd the epression π ( ) ( ) 0 for R is V = d Our region etends from = 0 to = V =.57 0
26 E 8 Let R e the region ounded = +, =, nd = 0. Find the volume of the solid generted when R is rotted out the is R r d c ( ) ( ) ( ) V = π R r d 0 = π 5 = 9.70 d 0
27 7. Homework Find the volume of the solid formed rotting the descried region out the given line.. =, =, = 0; out the is.. = e, = 0, =, = 0; out the is.. =, =, =, = 0; out the is.. =, = 0; out the is. 5. =, = 0, = ; out the is. 6. sec,, π π = = =, = ; out the is. 7. =, = ; out the is. 8. =, =, = 0; out the is. 9. =, = +, =, = ; out the is. 0. =, = ; out the is.. =, = e, = ; out the is.. =, = ; out the is. 0
28 7. Volume Rottion out Line not n Ais In the lst section we lerned how to find the volume of nonregulr solid. This section will e more of the sme, ut with it of twist. Insted of rotting out n is, let the region e revolved out line not the origin. Volume Disc Method (Form ): The volume of the solid generted when the f, is rotted out the line = k function f ( ) from = nd =, where ( ) 0 [or g( ) from = c nd = d, where g( ) 0, is rotted out the line given or ( ) or ( ) V = π f k d V = π d V = π g h d r d where r is the Length of our Riemnn rectngle. c = h is Volume Wsher Method (Prt ): The volume of the solid generted when g, from = nd =, the region ounded functions f ( ) nd ( ) where f ( ) g( ) [or f ( ) nd g( ), from = c nd = d, where f ( ) g( ) is rotted out the line = k is given π ( ) ( ) V = f k g k d d c or ( ) ( ) V = π f h g h d d ( ) V = π R r d c where R is the outer rdius of the Riemnn rectngle nd r is the inner rdius. ], 05
29 E Let R e the region ounded the equtions =, =, =, nd =. Find the volume of the solid generted when R is rotted out the line = r r ( ) = V = π r d 7 = π + = 58. d As in the previous section, this mkes n interesting shpe, ut is of ver little use to us in ctull solving for the volume. You could esil hve found the vlue for the volume never knowing wht the shpe ctull looked like. 06
30 z 07
31 π E Let R e the region ounded the curves = sec, =, =, nd 5π =. Find the volume of the solid generted when R is rotted out the line =. R π/ π π/ π 5π/ R= sec V = π d 5π sec π = π = 8.7 d 08
32 E Find the volume of the solid otined rotting out the is the region ounded = nd the line = out the line =. R ( ) ( ) R= d V = π + d c π = + d = Wht if ou were sked to tke the region ounded = nd = nd rotte it out the is? How is this different from the previous prolems? 09
33 E Let R e the region ounded the equtions =, the is, =, nd = 5. Find the volume of the solid generted when R is rotted out the line = R r 5 ( ) ( ) 0 ( ) V = π R r d ( ) 5 π = + =.78 d 0
34 π E 5 Let R e the region ounded the curves = sec, the is, =, 5π nd =. Find the volume of the solid generted when R is rotted out the line =. ( ) V = π R r d 5π ( sec ) ( π ) π = = 8.5 d R r π/ π π/ π 5π/ The rotted figure looks like this:
35 z It would e difficult to sketch this hnd, nd we definitel do not need it to solve the prolem.
36 E 6 Find the volume of the solid otined rotting out the is the region ounded = nd the is out line =. Given the informtion ove nd the sketch of = elow, sketch our oundries, the is of rottion, our Riemnn rectngle, nd our inner nd outer rdii. R r 5 6 Set up the integrtion: d c ( ) V = π R r d π 0 = 0.59 ( ) ( ) = + d
37 E 7 Let R e the region ounded the curves = sin, = cos, = 0, nd π =. Find the volume of the solid otined rotting R out the is π/ π/ π/ π/ π π c ( ) π ( ) V = π R r d + R r d π ( cos ) ( sin ) π ( sin ) ( cos π ) = π d + d = π 0
38 7. Homework Set A. Given the curves f ( ) = ln, g ( ) e = nd =. 5. Find the re of the region ounded three curves in the first qudrnt.. Find the volume of the solid generted rotting the region round the line =. c. Find the volume of the solid generted rotting the region round the line =. 5
39 . Let f nd g e the functions given f ( ) ( ) g( ) = ( ) = nd Find the re of the region ounded the two curves.. Find the volume of the solid generted rotting the region round the line =. c. Find the volume of the solid generted rotting the region round the line =. g = nd the d. Set up n integrl epression for the curve ( ) ( ) curve h ( ) ( ) = which is not pictured ove tht is rotted round the line = k, where k >, in which the volume equls 0. Do not solve this eqution. 6
40 Find the volume of the solid formed rotting the descried region out the given line. Sketch the grph so tht it is esier for ou to ppl the Disk/Wsher Methods.. =, = 0, =, = ; out the line =.. = + sin, =, = 0, = π; out the line =. 5. =, = ; out the line =. 6. =, = ; out the line =. Use our grpher to sketch the regions descried elow. Find the points of intersection nd find the volume of the solid formed rotting the descried region out the given line. 7. = ln( +), = cos; out the line =. 8. =, = ; out the line =. 9. = + sin, =, = 0, = π; out the line =. 0. =, =; out the line =. 7
41 7. Homework Set B f g = 6+ 5, find the re ounded the two curves in the first qudrnt.. Given the functions ( ) = + nd ( ) ( ). Find the volume of the solid generted when the function ln ( 9 ) rotted round the line = on the intervl, = + is. Find the volume of the solid generted when the re in the first qudrnt = ln + nd = ( ) is rotted round the ounded the curves ( ) is π. Let R e the region ounded the grphs sin = nd = ( 6) R Find the re of the region R.. Find the volume of the solid when the region R is rotted round the line = 0. c. The line = cuts the region R into two regions. Write nd evlute n integrl to find the re of the region elow the line. 8
42 8. Given the functions elow re = nd = +. Let A e the region ounded the two curves nd the is. A 5. Find the re of region A.. Find the vlue of d d for = + t the point where the two curves intersect. c. Find the volume of the solid when region A is rotted round the  is. 9
43 7. Volume the Shell Method There is nother method we cn use to find the volume of solid of rottion. With the Disc/Wsher method we drew our smple pieces perpendiculr to the is of rottion. Now we will look into clculting volumes when our smple pieces re drwn prllel to the is of rottion (prllel nd shell rhme tht s how ou cn rememer the go together). Ojectives: Find the volume of solid rotted when region is rotted out given line. Tke the following function, let s cll it ( ) f, from to nd rotte it out the is. 0
44 How do we egin to find the volume of this solid? Imgine tking tin strip of the function nd rotting this little piece round the is. Wht would tht piece look like once it ws rotted? Wht is the surfce re formul for clindricl shell? π rl Now would the surfce re of just this one piece give ou the volume of the entire solid? How would we dd up ll of the surfce res of ll these little pieces? We rrive t our volume formul. Volume Shell Method: The volume of the solid generted when the region f, from = nd =, where ( ) 0 g, from ounded function ( ) f [or ( ) = c nd = d, where g( ) 0], is rotted out the is is given V f ( ) d or V = π ( ) = π d c g d ** Everthing rhmes with the shell method Shell/Prllel/ π rl so the disc method must e of the form Disc/Perpendiculr/ π r.
45 E Let R e the region ounded the equtions =, = 0, =, =. Find the volume of the solid generted when R is rotted out the is V V 0 = π d Our pieces re prllel to the is, so our integrnd will contin d = π d We cnnot integrte with respect to so we will sustitute out for d d V = π The epression for is V = π Our region etends from = to = V = 0π
46 E Let R e the region ounded =, = 5, nd = 0. Find the volume of the solid generted when R is rotted out the is ( ) V = π 5 d Our pieces re prllel to the is, so our integrnd will contin d. We cnnot integrte with respect to so we will sustitute out for. V = π ( 5 ) d The epression for is. 5 ( ) V = π 5 d Our region etends from = 0 to = 5 0 V = 50π
47 E Let R e the region ounded =, =, = nd =. Find the volume of the solid generted when R is rotted out the is V d = π d c V = π d V = π d V = 0π
48 E Let R e the region ounded =, + =, nd = 0. Find the volume of the solid generted when R is rotted out the is. 5 V d = π d c Our pieces re prllel to the is, so our integrnd will contin d. But we will need two integrls s our region of rottion is ounded different curves. The dividing line is = π 0 V = π d+ d π 0 ( ) V = π d+ + d V = 7π The Shell Method cn e pplied to rotting regions out lines other thn the es, just s the Disk nd Wsher Methods were. 5
49 E 5 Let R e the region ounded the equtions =, the is, nd = 7. Find the volume of the solid generted when R is rotted out the line = d V = π c d Our pieces re prllel to the is, so our integrnd will contin d d ( ) ( ) V = π c 7 d We cnnot integrte with respect to so we will sustitute out for d ( ) ( ) ( ) ( ) ( ) ( ) V = π 7 + d The epression for is + c V = π 7 + d Our region etends from = 0 to 0 V = = 6
50 The Wsher Method ields the sme result. ( ) V = π R r d π ( ) ( ) V = d π 7 ( ) ( ) V = d V =
51 7. Homework Set A Find the volume of the solid formed rotting the descried region out the given line. π. = sec, =, = 0, = ; out the is. 6. =, = 0, = 8; out the is. =, =, = 8, = 0; out the is... = e, = 0, =, = ; out the is. = nd the is; out the is. 5. ( ) 6. = sin nd the is on 0, π ; out the is. Use our grpher to sketch the regions descried elow. Find the points of intersection nd find the volume of the solid formed rotting the descried region out the given line. 7. =, = e, = ; out the line =. 8. ( ) = ln +, = cos ; out the line =. 9. =, = ; out the line =. 8
52 7.5 Volume Cross Sections In the lst sections, we hve lerned how to find the volume of solid creted when region ws rotted out line. In this section, we will find volumes of solids tht re not mde though revolutions, ut cross sections. Volume Cross Sections Formul: The volume of solid mde of cross sections with re A is V = A( ) dor V ( ) = d c A d Ojectives: Find the volume of solid with given cross sections. Steps to Finding the Volume of Solid Cross Sections:. Drw picture of the region of the se.. Sketch smple cross section (the se of the smple cross section will lie on the se sketched in Step ) if the piece is verticl our integrl will hve d in it, if our piece is horizontl our integrl will hve d in it.. Determine n epression representing the re of the smple cross section (e creful when using semicircles the rdius is hlf the distnce of the se).. Determine the endpoints the region covers. 5. Set up n integrl contining the limits of integrtion found in Step 5, nd the integrnd epression found in Step. 9
53 E The se of solid S is + =. Cross sections perpendiculr to the is re squres. Wht is the volume of the solid? z View of the solid generted with the given crosssections from the perspective of the is coming out of the pge. 0
54 z View of the solid generted with the given crosssections from the perspective of the is coming out of the pge. V V V V ( ) = A d Strt here ( side) = d Are formul for our (squre) cross section ( ) = side d Endpoints of the region of our solid ( ) = side d Simplif the integrl 0 ( ) V = d Length of the side of our cross section V 0 = d We cnnot integrte with respect to so we will 0 0 ( ) V = d V =5. su out for
55 E The se of solid is the region S which is ounded the grphs π = sin nd = ( 6). Cross sections perpendiculr to the is re semicircles. Wht is the volume of the solid? It is esier for setup to visulize the se region nd the crosssection seprtel nd not to tr to imgine the D solid d 5 6 V ( ) = A d Strt here = π r d Are formul for our cross section V V = π r d Endpoints of the region of our solid 0 d V = π d 0 V = π d d 8 For our cross section = r. 0 ( ) π π V = sin d We cnnot integrte with respect to so we will sustitute out for. V = 0.08
56 Here is n illustrtion of the solid, cutw so tht ou cn see the crosssection: z d Two views of the solid from Emple z
57
58 E The se of solid is the region S which is ounded the grphs f ( ) = ( ) nd g( ) = ( ). Crosssections perpendiculr to the is re equilterl tringles. Wht is the volume of the solid? V = = 0 0 ( ) d d = d d 0 = ( ( ) ( ) ) d =.589 A d d 5
59 E Let R e the region in the first qudrnt ounded = cos, = e, nd π =. () Find the re of region R. () Find the volume of the solid otined when R is rotted out the line =. (c) The region R is the se of s solid. For this solid, ech cross section perpendiculr to the is is squre. Find the volume of the solid. π () = ( cos ) A e d=.80 0 () 5 R π/ π/ π/ ( ) π V = π R r d 0 π = π ( ) ( ) V d π 0 = ( ) ( cos ) = π + e + d 6
60 (c) 5 side π/ π/ V = π 0 π 0 side ( ) d ( cos ) V = e d V = 8.05 e cos e cos Wht ou m hve noticed is tht the se of our crosssection is lws either top curve ottom curve or right curve left curve. Wht this mens for us is tht ll we rell need to do is ) Drw the crosssection nd lel the se with its length (top ottom or right left). ) Figure out the re formul for tht crosssection. ) Integrte the re formul on the pproprite intervl. 7
61 7.5 Homework Find the volume of the solid descried.. The se is circle of rdius nd the crosssections re squres.. The se is the ellipse 9 + = 6 nd the crosssections perpendiculr to the is re isosceles right tringles with the hpotenuse in the se.. The se is the region nd the crosssections perpendiculr to the is re squres.. The se is the region nd the crosssections perpendiculr to the is re squres. 5. The se is the region nd the crosssections perpendiculr to the is re equilterl tringles. Use our grpher to sketch the regions descried elow. Find the points of intersection nd find the volume of the solid tht hs the descried region s its se nd the given crosssections. 6. =, = e, =; the crosssections re semicircles. 7. = ln( +) nd = cos; the crosssections re squres. 8. The infinite region ounded =, = in Qudrnt I where the cross sections re squres. 9. The crosssectionl res of felled tree cut into 0foot sections re given in tle elow. Use the Midpoint Riemnn Sum with n=5 to estimte the volume of the tree Are
62 8. Given the curves f ( ) = + +, ( ) 8 qudrnt. g = nd in the first S R T 5. Find the re of the regions R, S, nd T.. Find the volume of the solid generted rotting the curve g( ) = round the line = 8 on the intervl, c. Find the volume of the solid generted rotting the region S round the line =. d. Find the volume of the solid generted if R forms the se of solid whose cross sections re squres whose ses re perpendiculr to the is 9
63 9. Let f nd g e the functions given ( ) = + nd g ( ) = f e Find the re of the region ounded two curves ove.. Find the volume of the solid generted rotting the region round the line =. c. Find the volume of the solid generted rotting the re etween g nd the is round the line = 0. ( ) d. Find the volume of the solid generted if the region ove forms the se of solid whose cross sections re semicircles with ses perpendiculr to the is. 0
64 0. Let R nd S e the regions ounded the grphs ( ) = + = nd R S 5. Find the re of the regions R nd S.. Find the volume of the solid when the region S is rotted round the line =. c. Find the volume of the solid when the region R is rotted round the line =. d. Find the volume of the solid if the region S forms the se of solid whose cross sections re equilterl tringles with ses perpendiculr s to the is. (Hint: the re of n equilterl tringle is A = ).
65 . Let h e the function given h ( ) ln ( ) = + s grphed elow. Find ech of the following:. The volume of the solid generted when h is rotted round the line = on the intervl 0,. The volume of the solid formed if the region etween the curve nd the lines =, = 0, nd = forms the se of solid whose cross sections re rectngles with heights twice their se nd whose ses re perpendiculr to the is.
66 sin =. Let R e the region in the first qudrnt enclosed the is nd the grphs of f nd g, nd let S e the region in the first qudrnt enclosed the grphs of f nd g, s shown in the figure elow.. Let f nd g e the functions given f ( ) = + ( π ) nd g ( ) e R S. Find the re of R. Find the re of S c. Find the volume of the solid generted when S is revolved round the horizontl line = d. The region R forms the se of solid whose crosssections re rectngles with ses perpendiculr to the is nd heights equl to hlf the length of the se. Find the volume of the solid.
67 7.6 Arc Length Bck in our geometr ds ou lerned how to find the distnce etween two points on line. But wht if we wnt to find the distnce etween to points tht lie on nonliner curve? Ojectives: Find the rc length of function in Crtesin mode etween two points. Just s the re under curve cn e pproimted the sum of rectngles, rc length cn e pproimted the sum of eversmller hpotenuses. d +d d d Here is our rc length formul: Arc Length etween Two Points: Let f ( ) e differentile function such tht f '( ) of f ( ) over, is given d is continuous, the length L d d L= + d or L= + d d d c As with the volume prolems, we will use Mth 9 in lmost ll cses to clculte rc length.
68 E Find the length of the curve = + e from =0 to =. d L = d + Strt here d L = + ( e ) d Sustitute in for d d 0 L =9. 58 Mth 9 E Find the length of the curve = + from = to =. d L = + c d d ( ) d L = + + d L = E Use right hnd Riemnn sums with n = to estimte the rc length of = +ln from = to = 5. How does this pproimtion compre with the true rc length (use Mth 9)? Let f ( ) = + ( + ln ) Let ( ) f represent our integrnd 5 ( ln ) L + + d = f + f + f + f 5 ( ) ( ) ( ) ( ) = = Approimte using rectngles 5
69 Actul Vlue: 5 ( ) L= + + ln = 906. Note: The integrl is the rc length. Some people get confused s to how rectngle res cn get us length. Rememer tht the Riemnn rectngles re w of evluting n integrl pproimtion. Just like the re under the curve could e displcement if the curve ws velocit, in this cse, the re under the curve d + d is the length of the rc long the curve. d E Find the length of the curve f ( ) = t+ dt L = + d d d ( ) = + + d = 5. 5 on t 7. 6
70 7.6 Homework Set A Find the rc length of the curve.. = + 6 on 0,. = + ln on,. =. = ln sec ( ) on, 9 ( ) on 0, π 5. = ln on, 6. = e on 0, 7. Find the perimeter of ech of the two regions ounded = nd =. 8. Find the length of the rc long f ( ) cos 0 9. Find the length of the rc long ( ) π = t dt on 0,. f = t dt on,. 7
71 7.6 Homework Set B Find the lengths of the following curves on the given intervls:. = 5+ cos, 0, π. = e, 0, π =, 0,. ln( sec ). ( ) =,, 9 5. =, 0, 6. = 6 + 0,, Find the perimeter of the regions enclosed the two curves: 7. = + ln, = 8. = e, = 6 9. = 7+ 6, = 0. = 9, =, 0, Find the length of the following curves:. + =. + = 5 + = 6. ( ) 8
72 Volume Test. The re of the region enclosed = nd = is given () ( )d 0 () ( )d 0 (c) ( )d 0 (d) 0 ( )d (e) 0 ( )d. Which of the following integrls gives the length of the grph = tn etween = to = if 0 < < < π? () (c) + tn d () + tn d + sec d (d) + tn d (e) + sec d. Let R e the region in the first qudrnt ounded = e, = nd = ln. Wht is the volume of the solid generted when R is rotted out the  is? ().80 ().8 (c).86 (d).89 (e).9 9
73 . A region is ounded =, the is, the line = m, nd the line = m, where m > 0. A solid is formed revolving the region out the is. The volume of the solid () is independent of m. () (c) increses s m increses. decreses s m increses. (d) increses until m =, then decreses. (e) is none of the ove 5. Let R e the region in the first qudrnt ounded = sin, the is, nd = π. Which of the following integrls gives the volume of the solid generted when R is rotted out the is? π ( ) d () π d () π sin 0 0 ( ) d π (c) π sin π (d) π sin (e) π sin 0 0 ( ) d 0 ( ) d 6. The se of solid is the region enclosed = cos for π < < π. If ech crosssection of the solid perpendiculr to the is is squre, the volume of the solid is () π () π (c) π (d) π (e) 50
74 7. Let R e the region ounded the grphs ( ) ( ) sin g π = pictured elow. f = + nd () Find the re of R. () Find the volume of the figure if R is rotted round the line =. (c) The region, R, is the se of solid whose crosssections re squres perpendiculr to the is. Find the volume of this solid. f = + is rotted round the (d) Find the volume of the solid if onl ( ) is. 5
75 8. Let S e the region ounded the curves f ( ) = nd ( ) 5 g= () Find the re etween the two curves. () Find the volume formed rotting the region S round the line = 7. (c) Find the perimeter of the region S. (d) Find the volume formed rotting the region S round the  is. 5
76 7. Homework Homework. π π. π 5. π π 5 8. π 9. 59π 5 0. π Homework c c..690 ( ( ) ) 0 d. V = π k ( k ( ) ) d = π π π 5 5
77 7. Homework π Homework DNE R=.9, S= c d c d Homework
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