Area of a Surface of Revolution

AREA OF A SURFACE OF REVOLUTION

A surface of revolution is formed when a curve is rotated about a line. Such a surface is the lateral boundary of a solid of revolution of the type discussed in Sections 7.2 and 7.3. We want to define the area of a surface of revolution in such a way 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 cylinder with radius r and height h is taken to be A 2rh because we can imagine cutting the cylin- cut h der and unrolling it (as in Figure 1) to obtain a rectangle with dimensions 2r and h . r Likewise, we can take a circular cone with base radius r and slant height l , cut it along the dashed line in Figure 2, and flatten it to form a sector of a circle with radius l and cen- tral angle 2rl . We know that, in general, the area of a sector of a circle with radius 1 2 l and angle is 2 l (see Exercise 67 in Section 6.2) and so in this case it is h 2r A 1 l 2 1 l 2 rl 2πr 2 2 l FIGURE 1 Therefore, we define the lateral surface area of a cone to be A rl .

2πr

cut l

r ¨ l

FIGURE 2 What about more complicated surfaces of revolution? If we follow the strategy we used with arc length, we can approximate the original curve by a polygon. When this polygon is rotated about an axis, it creates a simpler surface whose surface area approximates the actual surface area. By taking a limit, we can determine the exact surface area. The approximating surface, then, consists of a number of bands, each formed by 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 3. The area of the band (or frus- l¡ tum of a cone) with slant height l and upper and lower radii r1 and r2 is found by sub- tracting the areas of two cones:

r¡ 1 A r2l1 l r1l1 r2 r1l1 r2l

l From similar triangles we have l1 l1 l r™ r1 r2 which gives FIGURE 3 r2l1 r1l1 r1l or r2 r1 l1 r1l Putting this in Equation 1, we get

A r1l r2l or

2 A 2rl

1 where r 2 r1 r2 is the average radius of the band.

y y=ƒ Now we apply this formula to our strategy. Consider the surface shown in Figure 4, which is obtained by rotating the curve y 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 x0, x1,..., xn and equal width x , as we 0 abx did in determining arc length. If yi f xi , then the point Pixi, yi lies on the curve. The

part of the surface between xi1 and xi is approximated by taking the line segment Pi1Pi and rotating it about the x -axis. The result is a band with slant height l Pi1Pi and aver- 1 (a) Surface of revolution age radius r 2 yi1 yi so, by Formula 2, its surface area is

P yi1 yi y i yi Pi-1 2 Pi1Pi P¸ Pn 2

As in the proof of Theorem 7.4.2, we have 0 abx

2 Pi1Pi s1 f xi* x

(b) Approximating band where xi* is some number in xi1, xi . When x is small, we have yi f xi f xi* and also yi1 f xi1f xi*, since f is continuous. Therefore FIGURE 4 y y i 1 i 2 2 Pi1Pi 2 f xi* s1 f xi* x 2

and so an approximation to what we think of as the area of the complete surface of revolution is

n 2 3 2 f xi* s1 f xi* x i1

This approximation appears to become better as n l and, recognizing (3) as a Riemann sum for the function tx 2 f x s1 f x2 , we have

n b 2 2 lim 2 f xi* s1 f xi* x y 2 f x s1 f x dx l n i1 a

Therefore, in the case where f is positive and has a continuous derivative, we deﬁne the surface area of the surface obtained by rotating the curve y f x ,a x b , about the x-axis as

b 4 S y 2 f x s1 f x2 dx a

With the Leibniz notation for derivatives, this formula becomes

2 b dy 5 S y 2y1 dx a dx

If the curve is described as x ty ,c y d , then the formula for surface area becomes

2 d dx 6 S y 2y1 dy c dy

length given in Section 7.4, as

7 S y 2y ds

For rotation about the y -axis, the surface area formula becomes

8 S y 2x ds

where, as before, we can use either

dy 2 dx 2 ds 1 dx or ds 1 dy dx dy

These formulas can be remembered by thinking of 2y or 2x as the circumference of a circle traced out by the point x, y on the curve as it is rotated about the x -axis or y -axis, respectively (see Figure 5).

y y

(x, y) y x (x, y) 0 x

circumference=2πy circumference=2πx 0 x

FIGURE 5 (a) Rotation about x-axis: S=j 2πy ds (b) Rotation about y-axis: S=j 2πx ds

EXAMPLE 1 The curvey s4 x 2 ,1 x 1 , is an arc of the circlex 2 y 2 4 . Find the area of the surface obtained by rotating this arc about the x -axis. (The surface is a portion of a sphere of radius 2. See Figure 6.)

y SOLUTION We have dy x 1 4 x 2 122x dx 2 s4 x 2

and so, by Formula 5, the surface area is

2 1 dy 1 x S y 2y 1 dx 1 dx

1 x 2 2 y s4 x 2 1 dx 1 4 x 2

1 2 2 y s4 x 2 dx 1 s x 2 FIGURE 6 4

2 ■■ Figure 7 shows the surface of revolution EXAMPLE 2 The arc of the parabola y x from 1, 1 to 2, 4 is rotated about the whose area is computed in Example 2. y-axis. Find the area of the resulting surface. y SOLUTION 1 Using

(2, 4) dy y x 2 and 2x dx y=≈ we have, from Formula 8,

0 12 x S y 2 x ds

2 FIGURE 7 2 dy y 2x 1 dx 1 dx

2 2 y x s1 4x 2 dx 1

Substituting u 1 4x 2 , we have du 8x dx . Remembering to change the limits of integration, we have

17 s 2 32 17 S y u du [ 3 u ]5 4 5 4 ■■ As a check on our answer to Example 2, (17s17 5s5) notice from Figure 7 that the surface area 6 should be close to that of a circular cylinder with the same height and radius halfway between SOLUTION 2 Using the upper and lower radius of the surface: 21.5328.27. We computed that dx 1 the surface area was x sy and dy 2sy (17s17 5s5) 30.85 6 we have which seems reasonable. Alternatively, the surface area should be slightly larger than the area 2 of a frustum of a cone with the same top and bot- 4 dx S y 2x ds y 2x 1 dy tom edges. From Equation 2, this is 1 dy 21.5(s10) 29.80.

4 1 4 2 y sy 1 dy y s4y 1 dy 1 4y 1

17 y su du (where u 1 4y ) 4 5 (17s17 5s5) (as in Solution 1) 6

EXAMPLE 3 Find the area of the surface generated by rotating the curve y e x , 0 x 1,about the x -axis.

we have

2 1 dy 1 S y 2y 1 dx 2 y e x s1 e 2x dx 0 dx 0

e 2 y s1 u 2 du (where u e x ) 1

2 y sec3 d (where u tan and tan1e ) 4

1 ■■ Or use Formula 21 in the Table of Integrals. 2 2 [sec tan ln sec tan ]4 (by Example 8 in Section 6.2) [sec tan lnsec tan s2 ln(s2 1)] Since tan e, we have sec2 1 tan2 1 e 2 and

S [es1 e 2 ln(e s1 e 2 ) s2 ln(s2 1)]

EXERCISES

17–20 Use Simpson’s Rule with n 10 to approximate the area A Click here for answers. S Click here for solutions. of the surface obtained by rotating the curve about the x -axis. Com- pare your answer with the value of the integral produced by your 1–4 Set up, but do not evaluate, an integral for the area of the calculator. surface obtained by rotating the curve about the given axis. 17. y ln x, 1 x 3 1. y ln x,;1 x 3 x-axis 18. y x sx, 1 x 2 2. y sin2x,;0 x 2 x-axis 19. y sec x, 0 x 3 3. y sec x,;0 x 4 y-axis 20. y s1 e x, 0 x 1 4. y e x,;1 y 2 y-axis ■■■■■■■■■■■■■

■■■■■■■■■■■■■

5–12 Find the area of the surface obtained by rotating the curve CAS 21–22 Use either a CAS or a table of integrals to ﬁnd the exact about the x -axis. area of the surface obtained by rotating the given curve about the x-axis. 5. y x 3, 0 x 2 21. y 1x, 1 x 2 6. 9x y 2 18, 2 x 6 22. y sx 2 1, 0 x 3 7. y sx, 4 x 9 ■■■■■■■■■■■■■ 8. y cos 2x, 0 x 6 CAS 23–24 Use a CAS to ﬁnd the exact area of the surface obtained 9. y cosh x, 0 x 1 by rotating the curve about the y -axis. If your CAS has trouble evaluating the integral, express the surface area as an integral in the x 3 1 10. y , 1 x 1 other variable. 6 2x 2 23. y x 3, 0 y 1 1 2 32 11. x 3 y 2 , 1 y 2 24. y lnx 1, 0 x 1 2 12. x 1 2y , 1 y 2 ■■■■■■■■■■■■■

■■■■■■■■■■■■■ 25. (a) If a 0 , ﬁnd the area of the surface generated by rotating 13–16 The given curve is rotated about the y -axis. Find the area the loop of the curve 3ay2 xa x2 about the x -axis. of the resulting surface. (b) Find the surface area if the loop is rotated about the 13. y s3 x, 1 y 2 y-axis. 14. y 1 x 2, 0 x 1 26. A group of engineers is building a parabolic satellite dish whose shape will be formed by rotating the curve y ax 2 x sa 2 y 2, 0 y a2 15. about the y -axis. If the dish is to have a 10-ft diameter and a 16. x a coshya, a y a maximum depth of 2 ft, ﬁnd the value of a and the surface area

27. The ellipse 32. Show that the surface area of a zone of a sphere that lies x 2 y 2 between two parallel planes is S dh , where d is the diam- 1 a b h a 2 b 2 eter of the sphere and is the distance between the planes. (Notice that S depends only on the distance between the planes is rotated about the x -axis to form a surface called an ellipsoid. and not on their location, provided that both planes intersect Find the surface area of this ellipsoid. the sphere.)

28. Find the surface area of the torus in Exercise 41 in Section 7.2. 33. Formula 4 is valid only when f x 0 . Show that when f x is not necessarily positive, the formula for surface area becomes 29. If the curve y f x ,a x b , is rotated about the horizon- tal line y c , where f x c , find a formula for the area of b S y 2 f x s1 f x2 dx the resulting surface. a CAS 30. Use the result of Exercise 29 to set up an integral to find the 34. Let L be the length of the curve y f x ,a x b , where area of the surface generated by rotating the curve y sx , f is positive and has a continuous derivative. Let S be the sur- 0 x 4,about the line y 4 . Then use a CAS to evaluate f face area generated by rotating the curve about the x -axis. If c the integral. is a positive constant, define tx f x c and let St be the 31. Find the area of the surface obtained by rotating the circle corresponding surface area generated by the curve y tx , 2 2 2 x y r about the line y r . a x b. Express St in terms of Sf and L . Thomson Brooks-Cole copyright 2007 AREA OF A SURFACE OF REVOLUTION ■ 7

ANSWERS

13. 145s145 10s10 27 15. a 2 S Click here for solutions. ( ) 17.9.023754 19. 13.527296 3 1. y 2 ln xs1 1x2 dx 21. 4 4 ln s17 4 4 ln s2 1 s17 4s2 1 [ ( ) ( ) ]

4 23. 6[ln(s10 3) 3s10] 3. y 2xs1 sec x tan x2 dx 0 25. (a) a 23 (b) 56s3a 245 5.(145s145 1)27 7. (37s37 17s17)6 27. 2[b 2 a 2b sin1(sa 2 b 2a)sa 2 b 2 ] 1 e 2 e2 xb s 2 2 2 9.[1 4 ] 11. 21 2 29.a 2 c f x 1 f x dx 31. 4 r Thomson Brooks-Cole copyright 2007 8 ■ AREA OF A SURFACE OF REVOLUTION

SOLUTIONS

1. y =lnx ds = 1+(dy/dx)2 dx = 1+(1/x)2 dx S = 3 2π(ln x) 1+(1/x)2 dx [by (7)] ⇒ ⇒ 1 3. y =secx ds = 1+(dy/dx)2 dx = 1+(secx tan x)2 dx ⇒ ⇒ π/4 2 S = 0 2πx 1+(secx tan x) dx [by (8)] 3 2 5. y = x y0 =3x .So ⇒ 2 2 2 3√ 4 4 3 S = 0 2πy 1+(y0) dx =2π 0 x 1+9x dx [u =1+9x , du =36x dx] 145 145 2π π 2 3/2 π √ = 36 1 √udu= 18 3 u = 27 145 145 1 1 − 7. y = √x 1+(dy/dx)2 =1+[1/(2√x )]2 =1+1/(4x).So ⇒

9 2 9 9 dy 1 1 S = 2πy 1+ dx = 2π √x 1+ dx =2π x + 4 dx 4 dx 4 4x 4 9 9 2 1 3/2 4π 1 3/2 π √ √ =2π 3 x + 4 = 3 8 (4x +1) = 6 37 37 17 17 4 4 − 9. y =coshx 1+(dy/dx)2 =1+sinh2 x =cosh2 x.So ⇒ S π 1 cosh x cosh xdx=2π 1 1 (1 + cosh 2x) dx = π x + 1 sinh 2x 1 =2 0 0 2 2 0 1 1 2 2 = π 1+ sinh 2 or π 1+ e e− 2 4 − 3/2 1/2 11. x = 1 y2 +2 dx/dy = 1 y2 +2 (2y)=y y2 +2 3 ⇒ 2 ⇒ 2 1+(dx/dy )2 =1+y2 y2 +2 = y2 +1 .So S =2π 2 y y2 +1 dy =2π 1 y4 + 1 y2 2 =2π 4+2 1 1 = 21π 1 4 2 1 − 4 − 2 2 13. y = √3 x x = y3 1+(dx/dy)2 =1+9y4.So ⇒ ⇒

2 2 2 3 4 2π 2 4 3 S =2π 1 x 1+(dx/dy) dy =2π 1 y 1+9y dy = 36 1 1+9y 36y dy

3/2 2 π 2 4 π √ √ = 18 3 1+9y = 27 145 145 10 10 1 −

2 2 1 2 2 1/2 2 2 15. x = a y dx/dy = (a y )− ( 2y)= y/ a y − ⇒ 2 − − − − ⇒ y2 a2 y2 y2 a2 1+(dx/dy)2 =1+ = − + = a2 y2 a2 y2 a2 y2 a2 y2 ⇒ − − − − a/2 a/2 a a/2 a S = 2π a2 y2 dy =2π ady =2πa y =2πa 0 = πa2. Note that this is 2 2 0 0 − a y 0 2 − − 1 1 4 the surface area of a sphere of radius a, and the length of the interval y =0to y = a/2 is 4 the length of the interval y = a to y = a. − Thomson Brooks-Cole copyright 2007 AREA OF A SURFACE OF REVOLUTION ■ 9

17. y =lnx dy/dx =1/x 1+(dy/dx)2 =1+1/x2 S = 3 2π ln x 1+1/x2 dx. ⇒ ⇒ ⇒ 1 2 3 1 1 Let f(x)=lnx 1+1/x .Sincen =10, ∆x = 10− = 5 . Then

1/5 S S10 =2π [f(1) + 4f(1.2) + 2f(1.4) + +2f(2.6) + 4f(2.8) + f(3)] 9.023754. ≈ · 3 ··· ≈ The value of the integral produced by a calculator is 9.024262 (to six decimal places).

19. y =secx dy/dx =secx tan x 1+(dy/dx)2 =1+sec2 x tan2 x ⇒ ⇒ ⇒ π/3 √ 2 2 √ 2 2 S = 0 2π sec x 1+sec x tan xdx.Letf(x)=secx 1+sec x tan x. π/3 0 π Since n =10, ∆x = − = .Then 10 30 π/30 π 2π 8π 9π π S S10 =2π f(0) + 4f +2f + +2f +4f + f 13.527296. ≈ · 3 30 30 ··· 30 30 3 ≈ The value of the integral produced by a calculator is 13.516987 (to six decimal places).

21. y =1/x ds = 1+(dy/dx)2 dx = 1+( 1/x2)2 dx = 1+1/x4 dx ⇒ − ⇒ 2 1 1 2 √x4 +1 4 √u2 +1 S = 2π 1+ dx =2π dx =2π 1 du [u = x2, du =2xdx] · x x4 x3 u2 2 1 1 1 4 4 √ 2 √ 2 1+u 24 1+u 2 = π 2 du = π +ln u + 1+u 1 u − u 1 = π √17 +ln 4+√17 + √2 ln 1+√2 = π √2 √17 +ln 4+√17 − 4 1 − − 4 1+√2

3 2 23. y = x and 0 y 1 y0 =3x and 0 x 1. ≤ ≤ ⇒ ≤ ≤

1 2 2 3 √ 2 1 2 S = 0 2πx 1+(3x ) dx =2π 0 1+u 6 du [u =3x , du =6xdx]

3 π 3 √ 2 21 π 1 √ 2 1 √ 2 = 3 0 1+u du = [or use CAS] 3 2 u 1+u + 2 ln u + 1+u 0

π 3 1 π = 3 2 √10 + 2 ln 3+√10 = 6 3 √10 + ln 3+√10

25. Since a>0,thecurve3ay2 = x(a x)2 only has points with − x 0. (3ay2 0 x(a x)2 0 x 0.) The ≥ ≥ ⇒ − ≥ ⇒ ≥ curve is symmetric about the x-axis (since the equation is

unchanged when y is replaced by y). y =0when x =0or a, − so the curve’s loop extends from x =0to x = a.

d d dy dy (a x)[ 2x + a x] (3ay2)= [x(a x)2] 6ay = x 2(a x)( 1) + (a x)2 = − − − dx dx − ⇒ dx · − − − ⇒ dx 6ay

dy 2 a2 6ax +9x2 12ax a2 6ax +9x2 a2 +6ax +9x2 (a +3x)2 1+ =1+ − = + − = = for x =0. dx 12ax 12ax 12ax 12ax 12ax 6 a a √x(a x) a +3x a (a x)(a +3x) (a) S = 2πy ds =2π − dx =2π − dx √3a · √12ax 6a x=0 0 0 a 2 π π a π π πa = (a2 +2ax 3x2) dx = a2x + ax2 x3 = (a3 + a3 a3)= a3 = . 3a − 3a − 0 3a − 3a · 3 0 Note that we have rotated the top half of the loop about the x-axis. This generates the full surface.

(b) We must rotate the full loop about the y-axis, so we get double the area obtained by rotating the top half of the loop:

a a a +3x 4π a S =2 2π xds=4π x dx = x1/2(a +3x) dx · √12ax 2 √3a x=0 0 0 2π a 2π 2 6 a 2π √3 2 6 = (ax1/2 +3x3/2) dx = ax3/2 + x5/2 = a5/2 + a5/2 √3a √3a 3 5 3 √a 3 5 0 0

2π √3 2 6 2π √3 28 56π √3 a2 = + a2 = a2 = 3 3 5 3 15 45

x2 y2 y (dy/dx) x dy b2x 27. + =1 = = a2 b2 ⇒ b2 −a2 ⇒ dx −a2y ⇒

dy 2 b4x2 b4x2 + a4y2 b4x2 + a4b2 1 x2/a2 a4b2 + b4x2 a2b2x2 1+ =1+ = = − = − dx a4y2 a4y2 a4b2 (1 x2/a2) a4b2 a2b2x2 − −

a4 + b2x2 a2x2 a4 a2 b2 x2 = − = − − a4 a2x2 a2(a2 x2) − −

The ellipsoid’s surface area is twice the area generated by rotating the ﬁrst quadrant portion of the ellipse about the

x-axis. Thus,

a dy 2 a b a4 (a2 b2)x2 S =2 2πy 1+ dx =4π a2 x2 − − dx dx a 2 2 0 0 − a √a x −

a a√a2 b2 4πb 4πb − du = a4 (a2 b2)x2 dx = a4 u2 [u = a2 b2 x] 2 2 2 2 a 0 − − a 0 − √a b − −

4 a√a2 b2 30 4πb u a 1 u − = a4 u2 + sin− 2 2 2 2 a √a b 2 − 2 a 0 − 2 2 2 1 √a b 2 2 4 2 2 a b sin− − 4πb a √a b a 1 √a b 2 a = − a4 a2(a2 b2)+ sin− − =2πb +  a2√a2 b2 2 − − 2 a √a2 b2 −  −     

29. The analogue of f(x∗) in the derivation of (4) is now c f(x∗),so i − i n b 2 2 S =lim 2π[c f(xi∗)] 1+[f 0(xi∗)] ∆x = 2π[c f(x)] 1+[f 0(x)] dx. n i=1 − − →∞ a Thomson Brooks-Cole copyright 2007 AREA OF A SURFACE OF REVOLUTION ■ 11

2 2 2 2 31. For the upper semicircle, f(x)=√r x , f 0(x)= x/√r x . The surface area generated is − − −

r 2 r 2 2 x 2 2 r S1 = 2π r r x 1+ dx =4π r r x dx 2 2 2 2 r − − r x 0 − − √r x − − − r r2 =4π r dx 2 2 0 √r x − −

r 2 2 2 x r For the lower semicircle, f(x)= √r x and f 0(x)= ,soS2 =4π + r dx. 2 2 2 2 − − √r x 0 √r x − − r 2 r r 2 1 x 2 π 2 2 Thus, the total area is S = S1 + S2 =8π dx =8π r sin− =8πr 2 =4π r . √r2 x2 r 0 0 − yi 1 + yi 33. In the derivation of (4), we computed a typical contribution to the surface area to be 2π − Pi 1Pi , the area 2 | − |

of a frustum of a cone. When f(x) is not necessarily positive, the approximations yi = f(xi) f(x∗) and ≈ i

yi 1 = f(xi 1) f(xi∗) must be replaced by yi = f(xi) f(xi∗) and yi 1 = f(xi 1) f(xi∗) . Thus, − − ≈ | | ≈ | | − | − | ≈ | |