# 6.2 Volumes of Revolution: The Disk Method

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1 mth ppliction: volumes of revolution, prt ii Volumes of Revolution: The Disk Method One of the simplest pplictions of integrtion (Theorem ) nd the ccumultion process is to determine so-clled volumes of revolution In this section we will concentrte on method known s the disk method Solids of Revolution If region in the plne is revolved bout line in the sme plne, the resulting object is solid of revolution, nd the line is clled the xis of revolution The following sitution is typicl of the problems we will encounter Solids of Revolution from Ares Under Curves Suppose tht y = f (x) is continuous (non-negtive) function on the intervl [, b] Rotte the region under the f between x = nd x = b round the the x-xis nd determine the volume of the resulting solid of revolution See Figure f (x i ) x i b y = f (x) f (x i ) x i b Once we know the cross-sectionl res of the solid, we cn use Theorem to determine the volume But s Figure shows, when the point (x i, f (x i )) on the curve is rotted bout the x-xis, it forms circulr cross-section of rdius R = f (x i ) Therefore, the cross-sectionl re t x i is Figure : Left: The region under the continuous curve y = f (x) on the intervl [, b] Right: The solid generted by rotting the region bout the x-xis Note: The point (x i, f (x i )) on the curve trces out circulr crosssection of rdius r = f (x i ) when rotted A(x i )=pr = p[ f (x i )] Since f is continuous, so is p[ f (x)] nd consequently Theorem pplies Volume of Solid of Revolution = A(x) dx = p[ f (x)] dx Of course, we could use this sme process if we rotted the region bout the y-xis nd integrted long the y-xis We gther these results together nd stte them s theorem THEOREM (The Disk Method) If V is the volume of the solid of revolution determined by rotting the continuous function f (x) on the intervl [, b] bout the x-xis, then V = p [ f (x)] dx () If V is the volume of the solid of revolution determined by rotting the continuous function f (y) on the intervl [c, d] bout the y-xis, then Z d V = p [ f (y)] dy () c Another Development of the Disk Method Using Riemnn Sums Insted of using Theorem, we could obtin Theorem directly by using the subdivide nd conquer strtegy once gin Since we will use this strtegy in lter situtions, let s quickly go through the rgument here There re often severl wys to prove result in mthemtics I hope one of these two will resonte with you

2 mth ppliction: volumes of revolution, prt ii 7 As bove, we strt with continuous function on [, b] This time, though, we crete regulr prtition of [, b] using n intervls nd drw the corresponding pproximting rectngles of equl width Dx In left hlf of Figure we hve drwn single representtive pproximting rectngle on the ith subintervl f (x i ) Dx b y = f (x) Dx b f (x i ) y = f (x) Rotting ech representtive rectngle cretes representtive disk (cylinder) of rdius R = f (x i ) (See the right hlf of Figure ) The volume of this cylinder is given by () Volume of Cylinder =(re of the bse) height In this cse when the disk is situted on its side, we think of the height s the width Dx of the disk Moreover, since the bse is circle, its re is pr = p[ f (x i )] so Volume of representtive disk = DV i = p[ f (x i )] Dx To determine the volume of entire solid of revolution, we tke ech pproximting rectngle, form the corresponding disks (see the middle pnel of Figure ) nd sum the resulting volumes, it genertes representtive disk whose volume is DV = pr Dx = p[r(x i )] Dx Figure : Left: The region under the continuous curve y = f (x) on the intervl [, b] nd representtive rectngle Right: The disk (cylinder) of rdius R = f (x i ) generted by rotting the representtive rectngle bout the x-xis The volume or this disk is pr w = p[ f (x i )] Dx Figure : A generl solid of revolution nd its pproximtion by series of n disks (Digrm from Lrson & Edwrds) Approximting the volume of the entire solid by n such disks (see the righthnd pnel of Figure ) of width Dx nd rdius f (x i ) produces Riemnn sum Volume of Revolution n Â i= p[ f (x i )] Dx = p n Â i= [ f (x i )] Dx () As usul, to improve the pproximtion we let the number of subdivisions n! nd tke limit Recll from our erlier work with Riemnn sums, this limit exists

3 mth ppliction: volumes of revolution, prt ii 8 becuse [ f (x)] is continuous on [, b] since f (x) is continuous there Volume of Revolution = lim n p n! Â i= [ f (x i )] Dx = p [ f (x)] dx (5) where we hve used the fct tht the limit of Riemnn sum is definite integrl This is the sme result we obtined in Theorem Ee could use this sme process if we rotted the region bout the y-xis nd integrted long the y-xis Stop! Notice how we used the subdivide nd conquer process to pproximte the quntity we wish to determine Tht is we hve subdivided the volume into pproximting disks whose volume we know how to compute We hve then refined this pproximtion by using finer nd finer subdivisions Tking the limit of this process provides the nswer to our question Identifying tht limit with n integrl mkes it possible to esily (!) compute the volume in question OK, time for some exmples I ll dmit it is hrd to drw figures like Figure However, drwing representtive rectngle for the region in question, s in the left hlf of Figure is usully sufficient to set up the required volume integrl Exmples Let s strt with couple of esy ones EXAMPLE Let y = f (x) =x on the intervl [, ] Rotte the region between the curve nd the x-xis round the x-xis nd find the volume of the resulting solid SOLUTION Using Theorem f (x) =x f (x) =x Wow, tht ws esy! Z V = p [ f (x)] dx = p [x ] dx = px5 5 = p 5 EXAMPLE Let y = f (x) =x on the intervl [, ] Rotte the region between the curve nd the y-xis round the y-xis nd find the volume of the resulting solid Figure : Left: A representtive rectngle for the curve y = x Right: A representtive circulr slice for the curve y = x rotted bout the x-xis f (x) =x SOLUTION The region is not the sme one s in Exmple It lies between the y-xis nd the curve, not the x-xis See Figure Since the rottion is bout the y-xis, we need to solve for x s function of y Since y = x, then x = p y Notice tht the region lies over the intervl [, ] on the y-xis now Using Theorem x = p y Z d Z V = p [g(y)] dy = p [ p y] dy = py c = p EXAMPLE 5 Find the volume of sphere of rdius r which cn be obtined by rotting the semi-circle f (x) = p r x bout the x-xis r r Figure 5: Left: A representtive rectngle for the curve y = p r x Right: A representtive circulr slice for the sphere tht results when rotting the semi-circle bout the x-xis Figure : Left: A representtive rectngle for the region between the curve y = x (x = p y) nd the y-xis Right: A representtive circulr slice for the curve x = p y rotted bout the x-xis

4 mth ppliction: volumes of revolution, prt ii 9 SOLUTION Using Theorem Z r V = p [ f (x)] dx = p [ p r x ] dx r Z r = p r x dx r = p r x r x r pple = p r r r + r = pr Amzing! We hve derived the volume formul of sphere from the volume by disks formul YOU TRY IT 9 Find the volume of cone of rdius r nd height h by rotting the line through (, ) nd (r, h) bout the y-xis (See Figure ) EXAMPLE (Two Pieces) Consider the region enclosed by the curves y = p x, y = x, nd the x-xis Rotte this region bout the x-xis nd find the resulting volume SOLUTION It is importnt to sketch the region to see the reltionship between the curves Py prticulr ttention to the bounding curves Drw representtive rectngles This will help you set up the pproprite integrls It is less importnt to try to drw very ccurte three-dimensionl picture h Figure : Determine the volume of cone of rdius r nd height h r First determine where the curves intersect: Obviously y = p x meets the x-xis t x = nd y = x meets the x-xis t x = For y = x nd y = p x, x = p x ) x + x = x ) x x + =(x )(x + 9) = ) x =, (x = 9 is not in the domin) From Figure 7 we see tht the solid is mde up of two seprte pieces (the top curve chnges t x = ) nd ech requires its own integrl Using Theorem Figure 7: Left: The region enclosed by the curves y = p x, y = x, nd the x-xis nd two representtive rectngles Right: The resulting solid of revolution bout the x-xis is formed by two distinct pieces ech requiring its own integrl Z V = p [ p Z x] dx + p [ x] dx Z Z = p xdx+ p [ x] dx = px p( x) + ( 8p) =(8p )+ = p Note: We used mentl djustment to do the second integrl (with u = x nd du = dx) Overll, the integrtion is esy once the problem is set up correctly Be sure you hve the correct region

5 mth ppliction: volumes of revolution, prt ii EXAMPLE 7 (One Piece: Two integrls) Reconsider the sme region s in Exmple enclosed by the curves y = p x, y = x, nd the x-xis Now rotte this region bout the y-xis insted nd find the resulting volume SOLUTION OK, the region is the sme s bove Here is where you hve to be very creful Since the rottion is bout the y-xis, the strips re horizontl this time Notice tht there re two strips When this region is rotted bout the y-xis, the solid will hve hollowed out center portion (see Figure 8) Thus, we must tke the outer region formed by the curve y = x nd subtrct the inner region formed by y = p x from it Since the rottion is bout the y-xis, to use the disk method we need to write the curves in the form x s function of y We hve Figure 8: Left: The region enclosed by the curves y = p x, y = x, nd the x-xis nd two representtive rectngles Right: The resulting solid of revolution bout the x-xis ( volcno ) is formed by two distinct pieces ech requiring its own integrl y = p x ) x = y nd y = x ) x = y The outer curve is x = y nd the inner curve is x = y from the perspective of the y-xis From the previous problem we lredy know the intersection points However we need their y-coordintes When x = the corresponding y-coordinte (using either y = p x or y = x) is y = The other coordinte is y = So this time using Theorem (note the subtrction of the inner volume) V = Outer Volume Inner Volume Z Z = p [ y] dy p [y ] dy p( y) py = 5 p p p = 5 = p 5 Whew! Tht s lot of things to process Getting cler picture of the region is crucil Even if you never drw the three-dimensionl version, understnding the representtive rectngles is criticl In this cse the representtive rectngles did not extend from the bottom to the top curve (ie, the left to the right), but rther from the y-xis to ech curve seprtely to represent the outer solid nd the inner solid Cution! We wrote the volume of the solid s the outer minus inner volume: Z Z V = p [ y] dy p [y ] dy Since the intervl is the sme for both integrls, we could hve written this s Z V = p [ y] [y ] dy where ech individul rdius is squred seprtely If you do combine the integrls, you cnnot squre the difference of the two rdii: Z V = p [ y y ] dy

6 mth ppliction: volumes of revolution, prt ii becuse [ y] [y ] =[ y y ] You ve been wrned! Look, there re two situtions you need to distinguish when multiple curves re used for solid of revolution The curves might crete two pieces which sum together to crete the entire solid in which cse you will need to dd integrls together to find the volume (see Figure 7) Or the curves might be situted so tht resulting solid is formed by hollowing out one solid from nother in which cse you will need to subtrct one integrl from nother (see Figure 8) This is why sketch of the position of the curves reltive to the xis of rottion is criticl Let s do couple more EXAMPLE 8 Consider the region enclosed by the curves y = p x, y = x x-xis Rotte this region bout the x-xis nd find the resulting volume, nd the SOLUTION Is this the sum of two integrls or is it difference of two integrls? First determine where the curves intersect: Obviously y = p x meets the x-xis t x = nd y = x meets the x-xis t x = Now y = x nd y = p x intersect when Outer Inner x = p x ) x x + = x ) x x + =(x )(x + ) = ) x =, (not x = ) From the sketch in Figure 9 if revolved bout the x-xis will result in solid tht hs been prtilly hollowed out ( cone hs been removed) This requires difference of integrls Using Theorem V = Outer Volume Inner Volume Z = p [ p Z x] dx p [x ] dx Z Z = p xdx p [x ] dx = px p(x ) 8p =(8p )+ = p EXAMPLE 9 Agin consider the region enclosed by the curves y = p x, y = x, nd the x-xis This time rotte the region bout the y-xis nd find the resulting volume Figure 9: Left: The region enclosed by the curves y = p x, y = x, nd the x-xis When rotted bout the x-xis, one region must be subtrcted from the other Insted of using representtive rectngles, we simply indicte the pproprite rdii with rrows SOLUTION Is this the sum of two integrls or is it difference of two integrls? Since the rottion is bout the y-xis, the rdii of the respective regions re horizontl, see Figure This is gin difference of two integrls Inner Trnslting the curves into functions of y we hve x = y, x = y +, nd y = (the Outer x-xis) The curves intersect the x-xis t y = We ve seen tht the line nd squre root function meet when x = since x = y + there, then the y-coordinte of the intersection is y = Figure : The region enclosed by the curves y = p x, y = x, nd the x-xis nd representtive rdii When rotted bout the y-xis, one region must be subtrcted from the other

7 mth ppliction: volumes of revolution, prt ii Using Theorem V = Outer Volume Z Z Inner Volume = p [y + ] dy p [y ] dy p(y + ) py = 5 p = = 8p 5 8p 5 p EXAMPLE Consider the region enclosed by the curves y = e x, x = nd the y-xis Rotte the region bout the y-xis nd find the resulting volume SOLUTION Is this the sum of two integrls or is it difference of two integrls? Since the rottion is bout the y-xis, the rdii of the respective regions re horizontl, see Figure This is gin difference of two integrls Trnslting the curves into functions of y we see y = e x becomes x = ln y Notice tht x = ln y meets the y-xis t nd it meets the line x = t y = e The solid formed by the region between the y-xis nd the curve x = ln y must be subtrcted from the ordinry cylinder formed by rotting the line x = bout the y-xis This cylinder hs rdius nd height e so its volume is v = p() e = pe (We could lso find this by integrting, but why bother) So Using Theorem Z e V = Outer Volume Inner Volume = pe p [ln y] dy =??? 5 e Inner Outer y = e x or x = ln y Figure : The region enclosed by the curves y = e x, x = nd the y-xis Rotte the region bout the y-xis nd find the resulting volume Since we don t know n ntiderivtive for (ln x) we re stuck (By the wy, notice tht the limits of integrtion for the integrl re from to e, not to e) So we will hve to invent nother method of finding such volumes if we re to solve this problem This shows the importnce of hving wide-vriety of methods for solving single problem YOU TRY IT Let R be the entire region enclosed by y = x nd y = x in the upper hlf-plne Sketch the region Rotte R bout the x-xis nd find the resulting volume (Answer: p) YOU TRY IT Let R be the region enclosed by y = x nd y = x () Rottion bout the x-xis Find the volume of the hollowed-out solid generted by revolving R bout the x-xis (Answer: p/5) (b) Rottion bout the y-xis Find the volume of the solid generted by revolving R bout the y-xis by using the disk method nd integrting long the y-xis (Answer: 8p/)

8 mth ppliction: volumes of revolution, prt ii Rottion About Other Axes It is reltively esy to dpt the disk method to finding volumes of solids of revolution using other horizontl or verticl xes The key steps re to determine the rdii of the slices nd express them in terms of the correct vrible A couple of exmples should give you the ide EXAMPLE Consider the region enclosed by the curves y = x x nd y = Rotte this region bout the line y = nd find the resulting volume SOLUTION The two curves meet when x x = ) x x =(x )(x + ) = ) x =, The prbol nd the line re esy to sketch; see Figure on the left y = y = x x y = A representtive rdius extends from the line y = to the curve y = x x The length of rdius is the difference between these two vlues, tht is, the rdius of circulr cross-section perpendiculr to the line y = is r = (x x) = + x x Since cross-section is circle, its re is A = A(x) =pr = p( + x x ) Since we know the cross-sectionl re, we cn use Theorem to find the volume Figure : Left: The region enclosed by the curves y = x x nd y = Since the xis of revolution is y =, representtive rdius extends from the line y = to the curve y = x x Right: The resulting solid of revolution bout the line y = V = Z A(x) dx = p ( + x x ) dx Z = p 9 + x x x + x ) dx = p 9x + x x x + x5 5 pple = p = 5p EXAMPLE Consider the region enclosed by the curves x = y nd x = y Rotte this region bout the line x = nd find the resulting volume SOLUTION The two curves meet when y = y ) y = ) y = ± The region is esy to sketch; see Figure on the left An outer representtive rdius extends from the line x = to the curve x = y nd n inner representtive rdius extends from the line x = to x = y The length of representtive rdius is the difference between the corresponding pirs of

9 mth ppliction: volumes of revolution, prt ii x = y x = y x = x = Figure : Left: The region enclosed by the curves x = y nd x = y nd two representtive rdii emnting from the xis of rottion x = Right: The resulting hollowed out solid of revolution bout the line x = looks like doughnut vlues The outer rdius is R = y nd the inner rdius is ( y )= + y The integrtion will tke plce long the y-xis on the intervl [, ] becuse the disks re horizontl when points re rotted bout the line x = Since we know the cross-sections re circles, we cn use Theorem to find the volume V = Outer Volume Z Z Inner Volume = p ( y ) dy p ( + y ) dy Z Z = p 9 y + y dy p + y + y dy Z = p 8 8y dy 8y = p 8y pple 8 = p = p Becuse the intervl of integrtion ws the sme for both the inner nd outer volumes, we re ble to combine the two integrls which gretly simplified the integrtion nd evlution If you understnd this problem, you re in good shpe YOU TRY IT Let R be the region enclosed by y = x, x =, nd the x-xis Drw the region () Rotte R bout the x-xis nd find the resulting volume (Answer: 8p/7) (b) Rotte R bout the line y = 9 nd find the resulting volume (Answer: 7p/7) (c) Rotte R bout the line x = nd find the resulting volume (Answer: p/5) YOU TRY IT Drw the region R in the first qudrnt enclosed by y = x, y = the x-xis x nd () Rotte R bout the x-xis nd find the resulting volume [Is it the sum of two integrls or outside minus inside?] (b) Rotte R bout the y-xis nd find the resulting volume [Is it the sum of two integrls or outside minus inside?] YOU TRY IT Let S be the region in the first qudrnt enclosed by y = x, y = x nd the y-xis () Rotte S bout the x-xis [Is it the sum of two integrls or outside minus inside?] (b) Rotte S bout the y-xis [Is it the sum of two integrls or outside minus inside?] YOU TRY IT 5 Let R be the region in the upper hlf-plne bounded by y = p x +, the x-xis nd the line y = x Find the volume resulting when R is rotted round the x-xis Remember: Outside minus inside How mny integrls do you need? YOU TRY IT A smll cnl bouy is formed by tking the region in the first qudrnt bounded by the y-xis, the prbol y = x, nd the line y = 5 x nd rotting it bout the y-xis (Units re feet) Find the volume of this bouy (Answer: p cubic feet) Figure : The region for you try it 5

10 mth ppliction: volumes of revolution, prt ii 5 YOU TRY IT 7 Just set up the integrls for ech of the following volume problems Simplify the integrnds where possible Use figures below () R is the region enclosed by y = x, y = x +, nd the y-xis in the first qudrnt Rotte R bout the x-xis (b) Sme R, but rotte round the y-xis (c) Sme R, but rotte round the line y = (d) S is the region enclosed by y = x + x +, y = x, the y-xis, nd the x-xis in the first qudrnt Rotte S bout the x-xis (e) T is the region enclosed by y = x + x +, y = x, nd the x-xis in the first qudrnt Rotte T bout the x-xis (f ) R is the region enclosed by y = x + nd y = x Rotte R bout the x-xis (g) Sme R, but rotte round the line y = (h) S is the region enclosed by y = x + nd y = x in the first qudrnt Rotte S bout the y-xis (i) T is the region enclosed by y = p x, y = x, the y-xis nd the x-xis Rotte T bout the y-xis (j) V is the region enclosed by y = p x, y = x, nd the x-xis Rotte V bout the y-xis (k) Sme V, but rotte round the x-xis (l) Hrd: Sme V, but rotte round the line y = c d e Figure 5: The regions for you try it 7 R S T i l f g h RR S T V SOLUTION Here re the solutions before the integnds hve been simplified Remember, you should simplify ech integrnd before ctully integrting Z () p Z (x + ) dx p Z (x ) dx (b) p ( p Z y) dy p (y ) dy Z Z Z Z (c) p ( x ) dx p ( (x + )) dx (d) p ( x + x + ) dx p (x ) dx Z Z (e) p (x ) dx + p ( x + x + ) dx (f ) Homework

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