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Unformatted text preview: 1134 C H A P T E R 17 LINE AND SURFACE INTEGRALS (ET CHAPTER 16) 43. Prove a famous result of Archimedes: The surface area of the portion of the sphere of radius r between two horizontal planes z = a and z = b is equal to the surface area of the corresponding portion of the circumscribed cylinder (Figure 22). a b z FIGURE 22 SOLUTION We compute the area of the portion of the sphere between the planes a and b . The portion S 1 of the sphere has the parametrization, 8( , ) = ( r cos sin , r sin sin , r cos ) where, D 1 : 2 , 1 If we assume 0 < a < b , then the angles and 1 are determined by, cos = b r = cos 1 b r cos 1 = a r 1 = cos 1 a r r b r a f f 1 The length of the normal vector is k n k = r 2 sin . We obtain the following integral: Area ( S 1 ) = ZZ D 1 k n k d d = Z 2 Z 1 r 2 sin d d = Z 2 r 2 d Z 2 1 sin d = 2 r 2 cos cos 1 a r = cos 1 b r = 2 r 2 a r + b r = 2 r ( b a ) The area of the part S 2 of the cylinder of radius r between the planes z = a and z = b is: Area ( S 2 ) = 2 r ( b a ) We see that the two areas are equal: Area ( S 1 ) = Area ( S 2 ) Further Insights and Challenges 44. Surfaces of Revolution Let S be the surface formed by revolving the region underneath the graph z = g ( y ) in the yz-plane for c y d about the z-axis (Figure 23). Assume that c 0. (a) Show that the circle generated by rotating a point ( , a , b ) about the z-axis is parametrized by ( a cos , a sin , b ), 2 S E C T I O N 17.4 Parametrized Surfaces and Surface Integrals (ET Section 16.4) 1135 (b) Show that S is parametrized by 8( y , ) = ( y cos , y sin , g ( y )) 12 for c y d , 0 2 . (c) Show that Area ( S ) = 2 Z d c y q 1 + g ( y ) 2 dy 13 z = g ( y ) (0, a , b ) ( a cos , a sin , 0) ( y cos , y sin , g ( y )) y x z a y x z ( a cos , a sin , b ) b (0, y , g ( y )) FIGURE 23 SOLUTION (a) The circle generated by rotating a point ( , a , b ) about the z-axis is a circle of radius a centered at the point ( , , b ) on the z-axis. Therefore it is parametrized by, ( a cos , a sin , b ), 2 (b) An arbitrary point ( x , y , g ( y )) on the surface S lies on the circle generated by rotating the point ( , y , g ( y )) about the z-axis. Using part (a), a parametrization of this circle is: ( y cos , y sin , g ( y )) , 2 Therefore, the following parametrization parametrizes the surface S : 8( y , ) = ( y cos , y sin , g ( y )) , 2 , c y d . (c) To compute the area of S we first find the tangent and normal vectors. We have: T y = 8 y = y ( y cos , y sin , g ( y )) =- cos , sin , g ( y ) T = 8 = ( y cos , y sin , g ( y )) = h y sin , y cos , i The normal vector is their cross product:...
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- Winter '08