AREA OF A SURFACE OF REVOLUTION

Transcription

1 AREA OF A SURFACE OF REVOLUTION h cut r πr h A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundar of a solid of revolution of the tpe discussed in Sections 7. and 7.. We want to define the area of a surface of revolution in such a wa that it corresponds to our intuition. If the surface area is A, we can imagine that painting the surface would require the same amount of paint as does a flat region with area A. Let s start with some simple surfaces. The lateral surface area of a circular clinder with radius r and height h is taken to be A rh because we can imagine cutting the clinder and unrolling it (as in Figure ) to obtain a rectangle with dimensions r and h. Likewise, we can take a circular cone with base radius r and slant height l, cut it along the dashed line in Figure, and flatten it to form a sector of a circle with radius l and central angle. We know that, in general, the area of a sector of a circle with radius l and angle is l (see Exercise 67 in Section 6.) and so in this case it is rl A l l r l rl FIGURE Therefore, we define the lateral surface area of a cone to be A rl. πr cut l r l FIGURE l What about more complicated surfaces of revolution? If we follow the strateg we used with arc length, we can approximate the original curve b a polgon. When this polgon is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area. B taking a limit, we can determine the exact surface area. The approximating surface, then, consists of a number of bands, each formed b rotating a line segment about an axis. To find the surface area, each of these bands can be considered a portion of a circular cone, as shown in Figure. The area of the band (or frustum of a cone) with slant height l and upper and lower radii r and r is found b subtracting the areas of two cones: r A r l l r l r r l r l r l From similar triangles we have l l l r r FIGURE which gives r l r l r l or r r l r l Putting this in Equation, we get or A r l r l Thomson Brooks-Cole copright 7 A rl where r r r is the average radius of the band.

2 AREA OF A SURFACE OF REVOLUTION =ƒ a b x (a) Surface of revolution Now we appl this formula to our strateg. Consider the surface shown in Figure, which is obtained b rotating the curve f x, a x b, about the x-axis, where f is positive and has a continuous derivative. In order to define its surface area, we divide the interval a, b into n subintervals with endpoints x, x,..., x n and equal width x, as we did in determining arc length. If i f x i, then the point P i x i, i lies on the curve. The part of the surface between x i and x i is approximated b taking the line segment P i P i and rotating it about the x-axis. The result is a band with slant height l P ip i and average radius r i i so, b Formula, its surface area is P i P i- P i P n i i P ip i a b x (b) Approximating band FIGURE As in the proof of Theorem 7.., we have where x i * is some number in x i, x i. When x is small, we have i f x i fx i * and also i f x i fx i *, since f is continuous. Therefore i i P ip i s f x i* x P ip i f x i * s f x i * x and so an approximation to what we think of as the area of the complete surface of revolution is n i f x i * s f x i * x This approximation appears to become better as n l and, recognizing () as a Riemann sum for the function tx f x s f x, we have lim n l n i f x i * s f x i * x b Therefore, in the case where f is positive and has a continuous derivative, we define the surface area of the surface obtained b rotating the curve f x, a x b, about the x-axis as a f x s f x S b a f x s f x With the Leibniz notation for derivatives, this formula becomes 5 S b a d If the curve is described as x t, c d, then the formula for surface area becomes Thomson Brooks-Cole copright 7 6 S d c d d and both Formulas 5 and 6 can be summarized smbolicall, using the notation for arc

3 AREA OF A SURFACE OF REVOLUTION length given in Section 7., as 7 S ds For rotation about the -axis, the surface area formula becomes 8 S x ds where, as before, we can use either ds d or ds d d These formulas can be remembered b thinking of or x as the circumference of a circle traced out b the point x, on the curve as it is rotated about the x-axis or -axis, respectivel (see Figure 5). (x, ) x x (x, ) FIGURE 5 circumference=π (a) Rotation about x-axis: S=j π ds circumference=πx x (b) Rotation about -axis: S=j πx ds EXAMPLE The curve s x, x, is an arc of the circle x. Find the area of the surface obtained b rotating this arc about the x-axis. (The surface is a portion of a sphere of radius. See Figure 6.) SOLUTION We have d x x x s x x and so, b Formula 5, the surface area is S d Thomson Brooks-Cole copright 7 FIGURE 6 Figure 6 shows the portion of the sphere whose surface area is computed in Example. s x x x s x s x 8

4 AREA OF A SURFACE OF REVOLUTION Figure 7 shows the surface of revolution whose area is computed in Example. EXAMPLE The arc of the parabola x from, to, is rotated about the -axis. Find the area of the resulting surface. SOLUTION Using (, ) = we have, from Formula 8, x and d x FIGURE 7 x S x ds x d x s x Substituting u x, we have du 8x. Remembering to change the limits of integration, we have S 7 5 su du [ u ] 5 7 As a check on our answer to Example, notice from Figure 7 that the surface area should be close to that of a circular clinder with the same height and radius halfwa between the upper and lower radius of the surface:. We computed that the surface area was (7s7 5s5).85 6 which seems reasonable. Alternativel, the surface area should be slightl larger than the area of a frustum of a cone with the same top and bottom edges. From Equation, this is..5(s) 9.8 SOLUTION we have Using S x ds 6 x s (7s7 5s5) and x d d d s s d s d su du (7s7 5s5) (where u ) (as in Solution ) Thomson Brooks-Cole copright 7 Another method: Use Formula 6 with x ln. EXAMPLE Find the area of the surface generated b rotating the curve e x, x,about the x-axis. SOLUTION Using Formula 5 with e x and d e x

5 AREA OF A SURFACE OF REVOLUTION 5 we have S d e x s e x e s u du sec d (where u e x ) tan e (where u tan and ) Or use Formula in the Table of Integrals. [sec tan ln sec tan ] (b Example 8 in Section 6.) [sec tan lnsec tan s ln(s )] Since tan e, we have sec tan e and S [es e ln(e s e ) s ln(s )] EXERCISES A Click here for answers. Set up, but do not evaluate, an integral for the area of the surface obtained b rotating the curve about the given axis.. ln x, x ; x-axis. sin x, x ; x-axis. sec x, x ; -axis. e x, ; -axis S Click here for solutions. 7 Use Simpson s Rule with n to approximate the area of the surface obtained b rotating the curve about the x-axis. Compare our answer with the value of the integral produced b our calculator. 7. ln x, 8. x sx, x x 9. sec x, x. s e x, x 5 Find the area of the surface obtained b rotating the curve about the x-axis. 5. x, x 6. 9x 8, 7. sx, 8. cos x, 9. cosh x,. x, 6 x x 9 x 6 x 6 x x. x,. x, CAS CAS Use either a CAS or a table of integrals to find the exact area of the surface obtained b rotating the given curve about the x-axis.. x,. sx, Use a CAS to find the exact area of the surface obtained b rotating the curve about the -axis. If our CAS has trouble evaluating the integral, express the surface area as an integral in the other variable.. x,. lnx, x x x Thomson Brooks-Cole copright 7 6 The given curve is rotated about the -axis. Find the area of the resulting surface.. s x,. x, x 5. x sa, a 6. x a cosha, a a 5. (a) If a, find the area of the surface generated b rotating the loop of the curve a xa x about the x-axis. (b) Find the surface area if the loop is rotated about the -axis. 6. A group of engineers is building a parabolic satellite dish whose shape will be formed b rotating the curve ax about the -axis. If the dish is to have a -ft diameter and a maximum depth of ft, find the value of a and the surface area of the dish.

6 6 AREA OF A SURFACE OF REVOLUTION CAS 7. The ellipse x a b a b is rotated about the x-axis to form a surface called an ellipsoid. Find the surface area of this ellipsoid. 8. Find the surface area of the torus in Exercise in Section If the curve f x, a x b, is rotated about the horizontal line c, where f x c, find a formula for the area of the resulting surface.. Use the result of Exercise 9 to set up an integral to find the area of the surface generated b rotating the curve sx, x,about the line. Then use a CAS to evaluate the integral.. Find the area of the surface obtained b rotating the circle x r about the line r.. Show that the surface area of a zone of a sphere that lies between two parallel planes is S dh, where d is the diameter of the sphere and h is the distance between the planes. (Notice that S depends onl on the distance between the planes and not on their location, provided that both planes intersect the sphere.). Formula is valid onl when f x. Show that when f x is not necessaril positive, the formula for surface area becomes S b a. Let L be the length of the curve f x, a x b, where f is positive and has a continuous derivative. Let S f be the surface area generated b rotating the curve about the x-axis. If c is a positive constant, define tx f x c and let S t be the corresponding surface area generated b the curve tx, a x b. Express in terms of and L. S t f x s f x S f Thomson Brooks-Cole copright 7

7 AREA OF A SURFACE OF REVOLUTION 7 ANSWERS.. S ln xs x Click here for solutions. xs sec x tan x 5. (5s5 )7 7. (7s7 7s7)6 9. [ e e ].. (5s5 s)7 5. a (a) a (b) 56sa 5 7. [ ln(s7 ) ln(s ) s7 s] 6[ln(s ) s] [b a b sin (sa b a)sa b ] 9. x b c f xs f x. r a Thomson Brooks-Cole copright 7

8 8 AREA OF A SURFACE OF REVOLUTION SOLUTIONS. =ln = +() = +() = (ln ) +() [b (7)]. = sec = +() = +(sec tan ) = +(sec tan ) [b (8)] 5. = =.So = +( ) = = 6 5 = 8 +9 [ =+9, =6 ] 5 = = +() =+[( )] =+(). So = 9 = + 9 = + 9 = 8 ( +) 9 + = = =cosh +() =+sinh =cosh.so = cosh cosh = ( + cosh ) = + sinh = + sinh or +. = + = + () = + +() =+ + = +.So = + = + = + =. = = +() =+9.So = +() = +9 = = 8 +9 = = = ( ) () = +() =+ = = + = = = = =.Notethatthisis Thomson Brooks-Cole copright 7 the surface area of a sphere of radius, andthelengthoftheinterval =to = is the length of the interval = to =.

9 AREA OF A SURFACE OF REVOLUTION 9 7. =ln = +() =+ = ln +. Let () =ln +.Since =, = = Then 5 = 5 [() + () + () + +(6) + (8) + ()] 975. The value of the integral produced b a calculator is 96 (to six decimal places). 9. =sec =sec tan +() =+sec tan = sec +sec tan.let() =sec +sec tan. Since =, = =.Then = () The value of the integral produced b a calculator is (to six decimal places) = = +() = +( ) = + = + = + + = [ =, =] + = + = +ln + + = 7 +ln ln = 7 +ln = and = and. = +( ) = + [ 6 =, =6] = = + = [or use CAS] + + ln ln + = 6 + ln + 5. Since,thecurve = ( ) onl has points with. ( ( ).) The curve is smmetric about the x-axis (since the equation is unchanged when is replaced b ). =when =or, so the curve s loop extends from =to =. Thomson Brooks-Cole copright 7 ( )= [( ) ] 6 = ( )() + ( ) ( )[ + ] = 6 = ( ) ( ) = ( ) ( ) 6 6 ( ) the last fraction is = ( )

10 AREA OF A SURFACE OF REVOLUTION + = = = ( +) = for 6=. (a) = = = = ( ) ( + ) = + ( )( +) = 6 + = ( + )= =. Note that we have rotated the top half of the loop about the x-axis. This generatesthefullsurface. (b) We must rotate the full loop about the -axis, so we get double the area obtained b rotating the top half of the loop: = = = + = ( +) = ( + ) = = = + 6 = 5 8 = () = = = + =+ = + = + = + ( ) = + = ( ) The ellipsoid s surface area is twice the area generated b rotating the first quadrant portion of the ellipse about the -axis. Thus, = + ( = ) = ( ) = [ = ] = + sin = ( )+ sin = + sin Thomson Brooks-Cole copright 7 9. The analogue of ( ) in the derivation of () is now ( ), so = lim = [ ( )] +[ ( )] = [ ()] +[ ()].

11 AREA OF A SURFACE OF REVOLUTION. For the upper semicircle, () =, () =.Thesurfaceareageneratedis = + = = For the lower semicircle, () = and () = Thus, the total area is = + =8,so = =8 sin. In the derivation of (), we computed a tpical contribution to the surface area to be + +. =8 =.,thearea of a frustum of a cone. When () is not necessaril positive, the approximations = ( ) ( ) and = ( ) ( ) must be replaced b = ( ) ( ) and = ( ) ( ). Thus, + ( ) +[ ( )]. Continuingwiththerestofthederivationasbefore,we obtain = () +[ ()]. Thomson Brooks-Cole copright 7