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 ෇ 2␲rh because we can imagine cutting the cylin- cut h der and unrolling it (as in Figure 1) to obtain a rectangle with dimensions 2␲r 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 ␪ ෇ 2␲r͞l. 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 2␲r A ෇ 1 l 2␪ ෇ 1 l 2ͩ ͪ ෇ ␲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 rotat- ing 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 ෇ ␲r2͑l1 ϩ l͒ Ϫ ␲r1l1 ෇ ␲͓͑r2 Ϫ r1͒l1 ϩ r2l͔ l From similar triangles we have ϩ l1 ෇ l1 l r™ r1 r2 which gives ෇ ϩ ͑ Ϫ ͒ ෇ FIGURE 3 r2l1 r1l1 r1l or r2 r1 l1 r1l Putting this in Equation 1, we get A ෇ ␲͑r1l ϩ r2l͒ or 2 A ෇ 2␲rl ෇ 1 ͑ ϩ ͒ where r 2 r1 r2 is the average radius of the band. 1 Thomson Brooks-Cole copyright 2007 2 ■ AREA OF A SURFACE OF REVOLUTION 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 ͑x͒,a ഛ x ഛ 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, b͔ 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 Pi͑xi, yi ͒ lies on the curve. The part of the surface between xiϪ1 and xi is approximated by taking the line segment PiϪ1Pi and rotating it about the x-axis. The result is a band with slant height l ෇ Խ PiϪ1Pi Խ and aver- ෇ 1 ͑ ϩ ͒ (a) Surface of revolution age radius r 2 yiϪ1 yi so, by Formula 2, its surface area is P yiϪ1 ϩ yi y i yi Pi-1 2␲ Խ PiϪ1Pi Խ P¸ Pn 2 As in the proof of Theorem 7.4.2, we have 0 abx 2 Խ PiϪ1Pi Խ ෇ s1 ϩ ͓ f Ј͑xi*͔͒ ⌬x ͓ ͔ ⌬ ෇ ͑ ͒Ϸ ͑ ͒ (b) Approximating band where xi* is some number in xiϪ1, xi . When x is small, we have yi f xi f xi* and also yiϪ1 ෇ f ͑xiϪ1͒Ϸf ͑xi*͒, since f is continuous. Therefore FIGURE 4 y Ϫ ϩ y i 1 i 2 2␲ Խ PiϪ1Pi Խ Ϸ 2␲ f ͑xi*͒ s1 ϩ ͓ f Ј͑xi*͔͒ ⌬x 2 and so an approximation to what we think of as the area of the complete surface of revo- lution is n 2 3 ͚ 2␲ f ͑xi*͒ s1 ϩ ͓ f Ј͑xi*͔͒ ⌬x i෇1 This approximation appears to become better as n l ϱ and, recognizing (3) as a Riemann sum for the function t͑x͒ ෇ 2␲ f ͑x͒ s1 ϩ ͓ f Ј͑x͔͒2, we have n b 2 2 lim ͚ 2␲ f ͑xi*͒ s1 ϩ ͓ f Ј͑xi*͔͒ ⌬x ෇ y 2␲ f ͑x͒ s1 ϩ ͓ f Ј͑x͔͒ dx l ϱ n i෇1 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 ͑x͒,a ഛ x ഛ b , about the x-axis as b 4 S ෇ y 2␲ f ͑x͒ s1 ϩ ͓ f Ј͑x͔͒2 dx a With the Leibniz notation for derivatives, this formula becomes 2 b dy 5 S ෇ y 2␲yͱ1 ϩ ͩ ͪ dx a dx If the curve is described as x ෇ t͑y͒,c ഛ y ഛ d , then the formula for surface area becomes 2 d dx 6 S ෇ y 2␲yͱ1 ϩ ͩ ͪ dy c dy and both Formulas 5 and 6 can be summarized symbolically, using the notation for arc Thomson Brooks-Cole copyright 2007 AREA OF A SURFACE OF REVOLUTION ■ 3 length given in Section 7.4, as 7 S ෇ y 2␲y ds For rotation about the y-axis, the surface area formula becomes 8 S ෇ y 2␲x ds where, as before, we can use either dy 2 dx 2 ds ෇ ͱ1 ϩ ͩ ͪ dx or ds ෇ ͱ1 ϩ ͩ ͪ dy dx dy These formulas can be remembered by thinking of 2␲y or 2␲x as the circumference of a circle traced out by the point ͑x, y͒ 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 Ϫ x 2 ,Ϫ1 ഛ x ഛ 1 , is an arc of the circlex 2 ϩ 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 Ϫx ෇ 1 ͑4 Ϫ x 2 ͒Ϫ1͞2͑Ϫ2x͒ ෇ dx 2 s4 Ϫ x 2 and so, by Formula 5, the surface area is 2 1 dy 1 x S ෇ y 2␲y ͱ1 ϩ ͩ ͪ dx Ϫ1 dx 1 x 2 ෇ 2␲ y s4 Ϫ x 2 ͱ1 ϩ dx Ϫ1 4 Ϫ x 2 1 2 ෇ 2␲ y s4 Ϫ x 2 dx Ϫ1 s Ϫ x 2 FIGURE 6 4 ■■ Figure 6 shows the portion of the sphere 1 ෇ 4␲ y 1 dx ෇ 4␲͑2͒ ෇ 8␲ whose surface area is computed in Example 1. Ϫ1 Thomson Brooks-Cole copyright 2007 4 ■ AREA OF A SURFACE OF REVOLUTION 2 ■■ Figure 7 shows the surface of revolution EXAMPLE 2 The arc of the parabola y ෇ x from ͑1, 1͒ to ͑2, 4͒ 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 2␲x ͱ1 ϩ ͩ ͪ dx 1 dx 2 2 ෇ 2␲ y x s1 ϩ 4x dx 1 Substituting u ෇ 1 ϩ 4x 2, we have du ෇ 8x dx. Remembering to change the limits of integration, we have ␲ ␲ ෇ 17 s ෇ 2 3͞2 17 S y u du [ 3 u ]5 4 5 4 ␲ ■■ As a check on our answer to Example 2, ෇ (17s17 Ϫ 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: 2␲͑1.5͒͑3͒Ϸ28.27. We computed that dx 1 the surface area was x ෇ sy and ෇ dy 2sy ␲ (17s17 Ϫ 5s5) Ϸ 30.85 6 we have which seems reasonable. Alternatively, the sur- face 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 2␲x ds ෇ y 2␲x ͱ1 ϩ ͩ ͪ dy tom edges. From Equation 2, this is 1 dy 2␲͑1.5͒(s10) Ϸ 29.80. 4 1 4 ෇ 2␲ y sy ͱ1 ϩ dy ෇ ␲ y s4y ϩ 1 dy 1 4y 1 ␲ 17 ෇ y su du (where u ෇ 1 ϩ 4y ) 4 5 ␲ ෇ (17s17 Ϫ 5s5) (as in Solution 1) 6 EXAMPLE 3 Find the area of the surface generated by rotating the curve y ෇ e x, 0 ഛ x ഛ 1,about the x-axis.