AREA OF A SURFACE OF REVOLUTION


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1 AREA OF A SURFACE OF REVOLUTION h cut r πr h A surfce of revolution is formed when curve is rotted bout line. Such surfce is the lterl boundr of solid of revolution of the tpe discussed in Sections 7. nd 7.. We wnt to define the re of surfce of revolution in such w 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 clinder with rdius r nd height h is tken to be A rh becuse we cn imgine cutting the clinder 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 Eercise 67 in Section 6.) nd so in this cse it is rl A l l r l rl FIGURE Therefore, we define the lterl surfce re of cone to be A rl. πr cut l r l FIGURE l Wht bout more complicted surfces of revolution? If we follow the strteg we used with rc length, we cn pproimte the originl curve b polgon. When this polgon is rotted bout n is, it cretes simpler surfce whose surfce re pproimtes the ctul surfce re. B tking limit, we cn determine the ect surfce re. The pproimting surfce, then, consists of number of bnds, ech formed b rotting line segment bout n is. 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 b subtrcting the res of two cones: r A r l l r l r r l r l r l From similr tringles we hve l l l r r FIGURE which gives r l r l r l or r r l r l Putting this in Eqution, we get or A r l r l Thomson BrooksCole copright 7 A rl where r r r is the verge rdius of the bnd.
2 AREA OF A SURFACE OF REVOLUTION =ƒ b () Surfce of revolution Now we ppl this formul to our strteg. Consider the surfce shown in Figure, which is obtined b rotting the curve f, b, bout the is, 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,,..., n nd equl width, s we did in determining rc length. If i f i, then the point P i i, i lies on the curve. The prt of the surfce between i nd i is pproimted b tking the line segment P i P i nd rotting it bout the is. The result is bnd with slnt height l P ip i nd verge rdius r i i so, b Formul, its surfce re is P i P i P i P n i i P ip i b (b) Approimting bnd FIGURE As in the proof of Theorem 7.., we hve P ip i s f i* where i * is some number in i, i. When is smll, we hve i f i f i * nd lso i f i f i *, since f is continuous. Therefore i i P ip i f i * s f i * nd so n pproimtion to wht we think of s the re of the complete surfce of revolution is n i f i * s f i * This pproimtion ppers to become better s n l nd, recognizing () s Riemnn sum for the function t f s f, we hve lim n l n i f i * s f i * b Therefore, in the cse where f is positive nd hs continuous derivtive, we define the surfce re of the surfce obtined b rotting the curve f, b, bout the is s f s f S b f s f With the Leibniz nottion for derivtives, this formul becomes 5 S b d If the curve is described s t, c d, then the formul for surfce re becomes Thomson BrooksCole copright 7 6 S d c d d nd both Formuls 5 nd 6 cn be summrized smbolicll, using the nottion for rc
3 AREA OF A SURFACE OF REVOLUTION length given in Section 7., s 7 S ds For rottion bout the is, the surfce re formul becomes 8 S ds where, s before, we cn use either ds d or ds d d These formuls cn be remembered b thinking of or s the circumference of circle trced out b the point, on the curve s it is rotted bout the is or is, respectivel (see Figure 5). (, ) (, ) FIGURE 5 circumference=π () Rottion bout is: S=j π ds circumference=π (b) Rottion bout is: S=j π ds EXAMPLE The curve s,, is n rc of the circle. Find the re of the surfce obtined b rotting this rc bout the is. (The surfce is portion of sphere of rdius. See Figure 6.) SOLUTION We hve d s nd so, b Formul 5, the surfce re is S d Thomson BrooksCole copright 7 FIGURE 6 Figure 6 shows the portion of the sphere whose surfce re is computed in Emple. s s s 8
4 AREA OF A SURFACE OF REVOLUTION Figure 7 shows the surfce of revolution whose re is computed in Emple. EXAMPLE The rc of the prbol from, to, is rotted bout the is. Find the re of the resulting surfce. SOLUTION Using (, ) = we hve, from Formul 8, nd d FIGURE 7 S ds d s Substituting u, we hve du 8. Remembering to chnge the limits of integrtion, we hve S 7 5 su du [ u ] 5 7 As check on our nswer to Emple, notice from Figure 7 tht the surfce re should be close to tht of circulr clinder with the sme height nd rdius hlfw between the upper nd lower rdius of the surfce:. We computed tht the surfce re ws (7s7 5s5).85 6 which seems resonble. Alterntivel, the surfce re should be slightl lrger thn the re of frustum of cone with the sme top nd bottom edges. From Eqution, this is..5(s) 9.8 SOLUTION we hve Using S ds 6 s (7s7 5s5) nd d d d s s d s d su du (7s7 5s5) (where u ) (s in Solution ) Thomson BrooksCole copright 7 Another method: Use Formul 6 with ln. EXAMPLE Find the re of the surfce generted b rotting the curve e,,bout the is. SOLUTION Using Formul 5 with e nd d e
5 AREA OF A SURFACE OF REVOLUTION 5 we hve S d e s e e s u du sec d (where u e ) tn e (where u tn nd ) Or use Formul in the Tble of Integrls. [sec tn ln sec tn ] (b Emple 8 in Section 6.) [sec tn lnsec tn s ln(s )] Since tn e, we hve sec tn e nd S [es e ln(e s e ) s ln(s )] EXERCISES A Click here for nswers. Set up, but do not evlute, n integrl for the re of the surfce obtined b rotting the curve bout the given is.. ln, ; is. sin, ; is. sec, ; is. e, ; is S Click here for solutions. 7 Use Simpson s Rule with n to pproimte the re of the surfce obtined b rotting the curve bout the is. Compre our nswer with the vlue of the integrl produced b our clcultor. 7. ln, 8. s, 9. sec,. s e, 5 Find the re of the surfce obtined b rotting the curve bout the is. 5., , 7. s, 8. cos, 9. cosh,., ,., CAS CAS Use either CAS or tble of integrls to find the ect re of the surfce obtined b rotting the given curve bout the is..,. s, Use CAS to find the ect re of the surfce obtined b rotting the curve bout the is. If our CAS hs trouble evluting the integrl, epress the surfce re s n integrl in the other vrible..,. ln, Thomson BrooksCole copright 7 6 The given curve is rotted bout the is. Find the re of the resulting surfce.. s,., 5. s, 6. cosh, 5. () If, find the re of the surfce generted b rotting the loop of the curve bout the is. (b) Find the surfce re if the loop is rotted bout the is. 6. A group of engineers is building prbolic stellite dish whose shpe will be formed b rotting the curve bout the is. If the dish is to hve ft dimeter nd mimum 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 b b is rotted bout the is to form surfce clled n ellipsoid. Find the surfce re of this ellipsoid. 8. Find the surfce re of the torus in Eercise in Section If the curve f, b, is rotted bout the horizontl line c, where f c, find formul for the re of the resulting surfce.. Use the result of Eercise 9 to set up n integrl to find the re of the surfce generted b rotting the curve s,,bout the line. Then use CAS to evlute the integrl.. Find the re of the surfce obtined b rotting the circle r bout the line 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 onl on the distnce between the plnes nd not on their loction, provided tht both plnes intersect the sphere.). Formul is vlid onl when f. Show tht when f is not necessril positive, the formul for surfce re becomes S b. Let L be the length of the curve f, b, where f is positive nd hs continuous derivtive. Let S f be the surfce re generted b rotting the curve bout the is. If c is positive constnt, define t f c nd let S t be the corresponding surfce re generted b the curve t, b. Epress in terms of nd L. S t f s f S f Thomson BrooksCole copright 7
7 AREA OF A SURFACE OF REVOLUTION 7 ANSWERS.. S ln s Click here for solutions. s sec tn 5. (5s5 )7 7. (7s7 7s7)6 9. [ e e ].. (5s5 s) [ ln(s7 ) ln(s ) s7 s] 6[ln(s ) s]. 5. () (b) 56s 5 7. [b b sin (s b )s b ] 9. b f s f. r Thomson BrooksCole copright 7
8 8 AREA OF A SURFACE OF REVOLUTION SOLUTIONS. =ln ds = +(d/) = +(/) S = π(ln ) +(/) [b (7)]. =sec ds = +(d/) = +(sec tn ) S = π/ π +(sec tn ) [b (8)] 5. = =.So S = π +( ) =π = π 6 5 udu= π 8 +9 [u =+9, du =6 ] 5 u/ = π = +(d/) =+[/( )] =+/(). So S = 9 =π π + d + / 9 = π 9 = π 8 ( +)/ 9 + =π = π =cosh +(d/) =+sinh =cosh.so S =π cosh cosh =π ( + cosh ) = π + sinh = π + sinh or π + e e. = + / /d = + / () = + +(/d) =+ + = +.So S =π + d =π + =π + = π. = = +(/d) =+9.So S =π +(/d) d =π +9 d = π +9 6 d 6 +9 / 5 5 = π 8 = π 7 5. = /d = ( ) / ( ) = / +(/d) =+ S = / = + = π / d =π d =π / =π = π. Note tht this is Thomson BrooksCole copright 7 the surfce re of sphere of rdius, nd the length of the intervl =to = / is the length of the intervl = to =.
9 AREA OF A SURFACE OF REVOLUTION 9 7. =ln d/ =/ +(d/) =+/ S = π ln +/. Let f() =ln +/.Sincen =, = = 5. Then S S =π /5 [f() + f(.) + f(.) + +f(.6) + f(.8) + f()] The vlue of the integrl produced b clcultor is 9.6 (to si deciml plces). 9. =sec d/ =sec tn +(d/) =+sec tn S = π/ π sec +sec tn.letf() =sec +sec tn. Since n =, = π/ = π.then S S =π π/ π f() + f +f π + +f 8π +f The vlue of the integrl produced b clcultor is (to si deciml plces). 9π π + f =/ ds = +(d/) = +( / ) = +/ S = π + =π + u + =π u du [u =, du =] +u = π du +u = π +ln u + +u u u = π 7 +ln ln = π 7 +ln = nd = nd. S = π +( ) =π +u du [u 6 =, du =6] = π +u du π = [or use CAS] u +u + ln u + +u = π + ln + = π 6 + ln + 5. Since >,thecurve = ( ) onl hs points with. ( ( ).) The curve is smmetric bout the is (since the eqution is unchnged when is replced b ). =when =or, so the curve s loop etends from =to =. Thomson BrooksCole copright 7 d ( )= d [( ) ] 6 d = ( )( ) + ( ) d ( )[ + ] = 6 d = ( ) ( ) = ( ) ( ) 6 6 ( ) the lst frction is / = ( )
10 AREA OF A SURFACE OF REVOLUTION + d = = = ( +) = for 6=. () S = = π = π ds =π ( + ) = π ( ) + ( )( +) =π 6 + = π ( + )= π = π. Note tht we hve rotted the top hlf of the loop bout the is. This genertes the full surfce. (b) We must rotte the full loop bout the is, so we get double the re obtined b rotting the top hlf of the loop: S = π ds=π = + = π / ( +) = π ( / + / ) = π / / = π 5/ / = π + 6 = π 5 8 = 56π (d/) = = d b b = b + d =+ b = b + = b + b / = b + b b b ( / ) b b = + b = b ( ) The ellipsoid s surfce re is twice the re generted b rotting the first qudrnt portion of the ellipse bout the is. Thus, S = π + d b ( =π b ) = πb ( b ) = πb b u du b [u = b ] = πb b u u + u b sin πb b = b ( b )+ b sin =π b + b sin b b Thomson BrooksCole copright 7 9. The nlogue of f( i ) in the derivtion of () is now c f( i ), so S = lim n n i= π[c f( i )] +[f ( i )] = b π[c f()] +[f ()].
11 AREA OF A SURFACE OF REVOLUTION. For the upper semicircle, f() = r, f () = / r. The surfce re generted is r S = π r r + =π r r =π r r r r For the lower semicircle, f() = r nd f () = Thus, the totl re is S = S + S =8π r r r r r,sos =π r r =8π r sin r r r r r r r + r. =8πr π =π r.. In the derivtion of (), we computed tpicl contribution to the surfce re to be π i + i P i P i, the re of frustum of cone. When f() is not necessril positive, the pproimtions i = f( i ) f( i ) nd i = f( i ) f( i ) must be replced b i = f( i ) f( i ) nd i = f( i ) f( i ). Thus, π i + i P i P i π f( i ) +[f ( i )]. Continuing with the rest of the derivtion s before, we obtin S = b π f() +[f ()]. Thomson BrooksCole copright 7
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