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1 AREA OF A SURFACE OF REVOLUTION cut h FIGURE FIGURE πr r r l h FIGURE A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundry of solid of revolution of the type discussed in Sections 7. nd 7.. We wnt to define the re of surfce of revolution in such wy tht it corresponds to our intuition. If the surfce re is A, we cn imgine tht pinting the surfce would require the sme mount of pint s does flt region with re A. Let s strt with some simple surfces. The lterl surfce re of circulr cylinder with rdius r nd height h is tken to be A rh becuse we cn imgine cutting the cylinder nd unrolling it (s in Figure ) to obtin rectngle with dimensions r nd h. Likewise, we cn tke circulr cone with bse rdius r nd slnt height l, cut it long the dshed line in Figure, nd fltten it to form sector of circle with rdius l nd centrl ngle. We know tht, in generl, the re of sector of circle with rdius l nd ngle is l (see Exercise 67 in Section 6.) nd so in this cse it is A l l r rl l Therefore, we define the lterl surfce re of cone to be A rl. Wht bout more complicted surfces of revolution? If we follow the strtegy we used with rc length, we cn pproximte the originl curve by polygon. When this polygon is rotted bout n xis, it cretes simpler surfce whose surfce re pproximtes the ctul surfce re. By tking limit, we cn determine the exct surfce re. The pproximting surfce, then, consists of number of bnds, ech formed by rotting line segment bout n xis. To find the surfce re, ech of these bnds cn be considered portion of circulr cone, s shown in Figure. The re of the bnd (or frustum of cone) with slnt height l nd upper nd lower rdii r nd r is found by subtrcting the res of two cones: From similr tringles we hve which gives rl cut Property of Cengge Lerning r l r l r l r l Putting this in Eqution, we get r l A r l l r l r r l r l l l l r r or πr r r l r l l or A r l r l Thomson BrooksCole copyright 7 A rl where r r r is the verge rdius of the bnd.
2 AREA OF A SURFACE OF REVOLUTION y y=ƒ b x () Surfce of revolution Now we pply this formul to our strtegy. Consider the surfce shown in Figure, which is obtined by rotting the curve y f x, x b, bout the xxis, where f is positive nd hs continuous derivtive. In order to define its surfce re, we divide the intervl, b into n subintervls with endpoints x, x,..., x n nd equl width x, s we did in determining rc length. If y i f x i, then the point P ix i, y i lies on the curve. The prt of the surfce between x i nd x i is pproximted by tking the line segment P ip i nd rotting it bout the xxis. The result is bnd with slnt height l PiPi nd verge rdius r y i y i so, by Formul, its surfce re is y FIGURE P P i P i P n b x (b) Approximting bnd y i As in the proof of Theorem 7.., we hve where x i * is some number in x i, x i. When x is smll, we hve y i f x ifx i * nd lso y i f x i fx i *, since f is continuous. Therefore nd so n pproximtion to wht we think of s the re of the complete surfce of revolution is This pproximtion ppers to become better s n l nd, recognizing () s Riemnn sum for the function tx f x s f x, we hve Therefore, in the cse where f is positive nd hs continuous derivtive, we define the surfce re of the surfce obtined by rotting the curve y f x, x b, bout the xxis s With the Leibniz nottion for derivtives, this formul becomes 5 lim n l n i yi yi PiPi s f xi * x P ip i n i f x i * s f x i * x y b S y b S y b f x i * s f x i * x yi yi PiPi f x i * s f x i * x f x s f x y dy f x s f x Property of Cengge Lerning If the curve is described s x ty, c y d, then the formul for surfce re becomes Thomson BrooksCole copyright 7 6 S y d c y dy dy nd both Formuls 5 nd 6 cn be summrized symboliclly, using the nottion for rc
3 AREA OF A SURFACE OF REVOLUTION length given in Section 7., s 7 S y y ds For rottion bout the yxis, the surfce re formul becomes y FIGURE 5 8 where, s before, we cn use either ds dy or These formuls cn be remembered by thinking of y or x s the circumference of circle trced out by the point x, y on the curve s it is rotted bout the xxis or yxis, respectively (see Figure 5). y EXAMPLE The curve y s x, x, is n rc of the circle x y. Find the re of the surfce obtined by rotting this rc bout the xxis. (The surfce is portion of sphere of rdius. See Figure 6.) SOLUTION () Rottion bout xxis: S=j πy ds We hve (x, y) circumference=πy dy x x nd so, by Formul 5, the surfce re is y S y x ds x S y y dy ds dy dy x s x y s x x x y (x, y) (b) Rottion bout yxis: S=j πx ds Property of Cengge Lerning x x circumference=πx x Thomson BrooksCole copyright 7 FIGURE 6 Figure 6 shows the portion of the sphere whose surfce re is computed in Exmple. y s x s x y 8
4 AREA OF A SURFACE OF REVOLUTION Figure 7 shows the surfce of revolution whose re is computed in Exmple. y EXAMPLE The rc of the prbol y x from, to, is rotted bout the yxis. Find the re of the resulting surfce. SOLUTION Using (, ) y= we hve, from Formul 8, y x nd dy x FIGURE 7 As check on our nswer to Exmple, notice from Figure 7 tht the surfce re should be close to tht of circulr cylinder with the sme height nd rdius hlfwy between the upper nd lower rdius of the surfce:. We computed tht the surfce re ws which seems resonble. Alterntively, the surfce re should be slightly lrger thn the re of frustum of cone with the sme top nd bottom edges. From Eqution, this is..5(s) 9.8 (7 s7 5 s5).85 6 x Substituting u x, we hve du 8x. Remembering to chnge the limits of integrtion, we hve SOLUTION we hve Using S y x ds y y7 5 su du (7s7 5s5) 6 S y x ds y x dy S y x s x x sy y7 5 su du (7s7 5s5) 6 nd x dy dy y sy y dy y sy dy (where u y ) (s in Solution ) [ u ] 5 7 dy sy Property of Cengge Lerning Thomson BrooksCole copyright 7 Another method: Use Formul 6 with x ln y. EXAMPLE Find the re of the surfce generted by rotting the curve y e x, x,bout the xxis. SOLUTION Using Formul 5 with y e x nd dy e x
5 AREA OF A SURFACE OF REVOLUTION 5 we hve S y y dy y e x s e x y e s u du y sec d (where u e x ) (where u tn nd ) tn e Or use Formul in the Tble of Integrls. A EXERCISES Click here for nswers. [sec tn ln sec tn ] [sec Set up, but do not evlute, n integrl for the re of the surfce obtined by rotting the curve bout the given xis.. y ln x, x ; xxis. y sin x, x ; xxis. y sec x, x ; yxis. y e x, y ; yxis 5 Find the re of the surfce obtined by rotting the curve bout the xxis. 5. y x, x 6. 9x y 8, 7. y sx, 8. y cos x, 9. y cosh x,. y x, 6 x x 9 x 6 x 6 x. x y, y. x y, y x S tn lnsec tn s ln(s )] Since tn e, we hve sec tn e nd Click here for solutions. (by Exmple 8 in Section 6.) S [es e ln(e s e ) s ln(s )] CAS CAS 7 Use Simpson s Rule with n to pproximte the re of the surfce obtined by rotting the curve bout the xxis. Compre your nswer with the vlue of the integrl produced by your clcultor. 7. y ln x, 8. y x sx, 9. y sec x, x. y s e x, x Use either CAS or tble of integrls to find the exct re of the surfce obtined by rotting the given curve bout the xxis.. y x,. y sx, Use CAS to find the exct re of the surfce obtined by rotting the curve bout the yxis. If your CAS hs trouble evluting the integrl, express the surfce re s n integrl in the other vrible.. y x, y. y lnx, x x x x Property of Cengge Lerning x Thomson BrooksCole copyright 7 6 The given curve is rotted bout the yxis. Find the re of the resulting surfce.. y s x, y. y x, x 5. x s y, y 6. x coshy, y 5. () If, find the re of the surfce generted by rotting the loop of the curve y x x bout the xxis. (b) Find the surfce re if the loop is rotted bout the yxis. 6. A group of engineers is building prbolic stellite dish whose shpe will be formed by rotting the curve y x bout the yxis. If the dish is to hve ft dimeter nd mximum depth of ft, find the vlue of nd the surfce re of the dish.
6 6 AREA OF A SURFACE OF REVOLUTION CAS 7. The ellipse x y b b is rotted bout the xxis to form surfce clled n ellipsoid. Find the surfce re of this ellipsoid. 8. Find the surfce re of the torus in Exercise in Section If the curve y f x, x b, is rotted bout the horizontl line y c, where f x c, find formul for the re of the resulting surfce.. Use the result of Exercise 9 to set up n integrl to find the re of the surfce generted by rotting the curve y sx, x,bout the line y. Then use CAS to evlute the integrl.. Find the re of the surfce obtined by rotting the circle x y r bout the line y r.. Show tht the surfce re of zone of sphere tht lies between two prllel plnes is S dh, where d is the dimeter of the sphere nd h is the distnce between the plnes. (Notice tht S depends only on the distnce between the plnes nd not on their loction, provided tht both plnes intersect the sphere.). Formul is vlid only when f x. Show tht when f x is not necessrily positive, the formul for surfce re becomes S y b. Let L be the length of the curve y f x, x b, where f is positive nd hs continuous derivtive. Let S f be the surfce re generted by rotting the curve bout the xxis. If c is positive constnt, define tx f x c nd let S t be the corresponding surfce re generted by the curve y tx, x b. Express in terms of nd L. S t f x s f x Property of Cengge Lerning S f Thomson BrooksCole copyright 7
7 AREA OF A SURFACE OF REVOLUTION 7.. S y y ANSWERS Click here for solutions. ln xs x xs sec x tn x 5. (5s5 )7 7. (7s7 7s7)6 9. [ e e ].. (5s5 s) () (b) 56s 5 7. [b b sin (s b )s b ] 9. x b f xs f x. r [ ln(s7 ) ln(s ) s7 s] 6[ln(s ) s] Property of Cengge Lerning Thomson BrooksCole copyright 7
8 8 AREA OF A SURFACE OF REVOLUTION SOLUTIONS. y =lnx ds = +(dy/) = +(/x) S = π(ln x) +(/x) [by (7)]. y =sin x ds = +(dy/) = +(sinx cos x) S = π/ π sin x +(sinx cos x) [by (7)]. y =secx ds = +(dy/) = +(secx tn x) S = π/ πx +(secx tn x) [by (8)]. y = e x ds = +(dy/) = +e x S = ln πx +e x [by (8)] or π(ln y) +(/y) dy [by (6)] 5. y = x y =x.so S = πy +(y ) =π x +9x [u =+9x, du =6x ] 5 udu= π 5 5 = π u/ = π 7 6. The curve 9x = y +8is symmetric bout the xxis, so we only use its top hlf, given by y = x. dy/ = 9 x,so+(dy/) =+ (x ). Thus, 6 S = π x + =6π x + 9 (x ) =6π x + / x =6π + / 6 =π 5 / 9 / =π 5 7. y = x +(dy/) =+[/( x )] =+/(x). So S = 9 =π dy 9 πy + = π x + =π x =π 98 =9π 8 Property of Cengge Lerning x + / 9 = π 9 (x +)/ 9 = π x + 8. y =cosx ds = +(dy/) = +( sinx) Thomson BrooksCole copyright 7 S = π/6 π cos x +sin x=π +u du [u =sinx, du =cosx] = π u +u + ln u + +u = π + ln + = π + π ln +
9 AREA OF A SURFACE OF REVOLUTION 9 9. y =coshx +(dy/) =+sinh x =cosh x.so S =π cosh x cosh x=π ( + cosh x) = π x + sinh x = π + sinh or π + e e. y = x 6 + x dy = x x x +(dy/) = + + x x = + = x x + x x S = π / 6 + x x + x 5 =π x / + x + x + x x 5 =π / + x + x x 6 =π 7 + x 6 x 8 / =π =π 6 = 6. x = y + / 5 56 π /dy = y + / (y) =y y + +(/dy) =+y y + = y +.So S =π y y + dy =π y + y =π + = π. x =+y +(/dy) =+(y) =+6y.So S =π y +6y dy = π = π. y = x x = y +(/dy) =+9y.So 6y + / ydy= π 6 6y + / S =π x +(/dy) dy =π y +9y dy = π 6 +9y / 5 5 = π 8 = π 7. y = x +(dy/) =+x S =π x +x = π 8x x + = π x + / +9y 6y dy Property of Cengge Lerning = π x = y /dy = ( y ) / ( y) = y/ y +(/dy) =+ S = / y y = y y + y y = y π / y y dy =π dy =π y / =π = π. Note tht this is Thomson BrooksCole copyright 7 the surfce re of sphere of rdius, nd the length of the intervl y =to y = / is the length of the intervl y = to y =.
10 AREA OF A SURFACE OF REVOLUTION 6. x = cosh(y/) +(/dy) =+sinh (y/) =cosh (y/).so y y S =π cosh cosh dy =π cosh y dy =π +cosh y dy =π y + y sinh =π + sinh =π + sinh or π e + e 7. y =lnx dy/ =/x +(dy/) =+/x S = π ln x +/x. Let f(x) =lnx +/x.sincen =, x = =. Then 5 S S =π /5 [f() + f(.) + f(.) + +f(.6) + f(.8) + f()] The vlue of the integrl produced by clcultor is 9.6 (to six deciml plces). 8. y = x + x dy/ =+ x / +(dy/) =+x / + x S = π x + x + + x x. Letf(x) =(x + x ) + + x x. Since n =, x = =.Then S S =π / [f() + f(.) + f(.) + +f(.8) + f(.9) + f()] The vlue of the integrl produced by clcultor is (to six deciml plces). 9. y =secx dy/ =secx tn x +(dy/) =+sec x tn x S = π/ π sec x +sec x tn x.letf(x) =secx +sec x tn x. Since n =, x = π/ = π.then S S =π π/ π π 8π 9π π f() + f +f + +f +f + f The vlue of the integrl produced by clcultor is (to six deciml plces).. y =(+e x ) / dy = ( + ex ) / e x = ( + e x ) / + S = dy Property of Cengge Lerning =+ e x e x ( + e x ) = +ex + e x = (ex +) ( + e x ) ( + e x ) π +e x e x + +e x = π (e x +) = π e x +x Let f(x) = (ex +).Sincen =, x = =.Then = π[(e +) ( + )] = π(e +). Thomson BrooksCole copyright 7 S S =π / [f() + f(.) + f(.) + +f(.8) + f(.9) + f()].68. The vlue of the integrl produced by clcultor is.687 (to six deciml plces).
11 AREA OF A SURFACE OF REVOLUTION. y =/x ds = +(dy/) = +( /x ) = +/x S = π + x x =π x + u + =π x u du [u = x, du =x] +u = π du +u = π +ln u + +u u u = π 7 +ln ln = π 7 +ln. y = x + dy = x x + S = π x + + x =π x dy ds = + = + x x + x + = π x + = π x x + x + ln + x + = π 9+ + ln + 9+ ln = 9 π + + ln = π 9 + ln + 9 = 9π + π ln + 9. y = x nd y y =x nd x. S = πx +(x ) =π +u du [u 6 =x, du =6x] +u du = π [or use CAS] u +u + ln u + +u = π = π + ln + = π 6 + ln + dy. y =ln(x +), x. ds = + = +,so x + S = =π πx + (x +) = +u u u π(u ) + du [u = x +, du = ] u +u du π du =π u 9 +u +u du π du u + ln Property of Cengge Lerning Thomson BrooksCole copyright 7, = [or use CAS] π u +u + u ln + +u + +u +u π ln u =π 5+ + ln 5 + ln π 5 ln + 5 +ln + =π ln + 5 +ln ln
12 AREA OF A SURFACE OF REVOLUTION 5. Since >,thecurvey = x( x) only hs points with x. (y x( x) x.) The curve is symmetric bout the xxis (since the eqution is unchnged when y is replced by y). y =when x =or, so the curve s loop extends from x =to x =. d (y )= d [x( x) ] 6y dy = x ( x)( ) + ( x) dy ( x)[ x + x] = 6y dy = ( x) ( x) = ( x) ( x) 6 y 6 x( x) the lst frction is /y = ( x) x dy + =+ 6x +9x = x x x + 6x +9x = +6x +9x ( +x) = for x 6=. x x x () S = = π x= πy ds =π x( x) +x =π x ( x)( +x) 6 ( +x x ) = π x + x x = π ( + )= π = π. Note tht we hve rotted the top hlf of the loop bout the xxis. This genertes the full surfce. (b) We must rotte the full loop bout the yxis, so we get double the re obtined by rotting the top hlf of the loop: S = π x= xds=π x +x = π x x / ( +x) = π (x / +x / ) = π x/ x5/ = π 5/ / = π + 6 = π 8 = 56π In generl, if the prbol y = x, c x c, is rotted bout the yxis, the surfce re it genertes is π Property of Cengge Lerning c x +(x) =π = π c c u +u du u =x, du = +u / π udu= +u / c = π 6 + c / Thomson BrooksCole copyright 7 Here c =ft nd c =ft, so c =5nd =. Thus, the surfce re is 5 S = π / 65 = 5π 5 9. ft + = 65π 6 / 5 = 65π 5
13 AREA OF A SURFACE OF REVOLUTION 7. x + y y (dy/) = = x dy b b = b x y + dy =+ b x y = b x + y = b x + b x / = b + b x b x y b ( x / ) b b x = + b x x x = b x ( x ) Theellipsoid ssurfcereistwicetheregenertedbyrottingthefirst qudrnt portion of the ellipse bout the xxis. Thus, S = = πb dy b ( πy + =π x b )x x ( b )x = πb πb u = u b + u sin b b u du b [u = b x] πb b = b ( b )+ b sin =π b + b sin b b 8. The upper hlf of the torus is generted by rotting the curve (x R) + y = r, y>, bout the yxis. y dy dy (x R) = (x R) + =+ = y +(x R) r = y y r (x R).Thus, S = =πr =πr R+r R r r r r r =πr +8πRr dy R+r rx πx + =π r (x R) u + R du [u = x R] r u udu r r u +πrr r du r u r R r du r u =8πRr sin (u/r) r =8πRr π =π Rr 9. Thenlogueoff(x i ) in the derivtion of () is now c f(x i ), so S = lim Property of Cengge Lerning n n i= π[c f(x i )] +[f (x i )] x = [since the first integrnd is odd nd the second is even] b π[c f(x)] +[f (x)]. Thomson BrooksCole copyright 7. y = x / y = x / +(y ) =+/x, sobyexercise9, S = π( x ) +/(x). UsingCAS,wegetS =π ln π
14 AREA OF A SURFACE OF REVOLUTION. For the upper semicircle, f(x) = r x, f (x) = x/ r x. The surfce re generted is r S = π r r x + x =π r r x =π r r r x r r r r r x r x For the lower semicircle, f(x) = r x nd f x r (x) = r x,sos =π r r x + r. Thus, the totl re is S = S + S =8π r. Tke the sphere x + y + z = d nd let the intersecting plnes be y = c nd y = c + h,where d c d h. Thesphereintersectsthexyplne in the circle x + y = d. From this eqution, we get x dy + y =, so dy = y. The desired surfce re is x r =8π r sin x r r x r =8πr π =π r. S =π xds=π c+h x +(/dy) dy =π c+h x +y /x dy =π c+h x + y dy c =π c+h ddy = πd c+h dy = πdh c c. In the derivtion of (), we computed typicl contribution to the surfce re to be π c c yi + yi P i P i,there of frustum of cone. When f(x) is not necessrily positive, the pproximtions y i = f(x i) f(x i ) nd y i = f(x i ) f(x i ) must be replced by y i = f(x i) f(x i ) nd y i = f(x i ) f(x i ). Thus, π yi + yi P i P i π f(x i ) obtin S = b +[f π f(x) (x)].. Since g(x) =f(x)+c,wehveg (x) =f (x). Thus, S g = b +[f (x i )] x. Continuing with the rest of the derivtion s before, we Property of Cengge Lerning = b πg(x) +[g (x)] = b πf(x) +[f (x)] +πc π[f(x)+c] +[f (x)] b +[f (x)] = S f +πcl Thomson BrooksCole copyright 7
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