17.4 Ex 43-51

# 17.4 Ex 43-51 - 1134 C H A P T E R 17 LINE AND SURFACE...

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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( θ , φ ) = ( r cos θ sin φ , r sin θ sin φ , r cos φ ) where, D 1 : ≤ θ ≤ 2 π , φ ≤ φ ≤ φ 1 If we assume 0 < a < b , then the angles φ and φ 1 are determined by, cos φ = b r ⇒ φ = cos − 1 b r cos φ 1 = a r ⇒ φ 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 φ . We obtain the following integral: Area ( S 1 ) = ZZ D 1 k n k d φ d θ = Z 2 π Z φ 1 φ r 2 sin φ d φ d θ = Ã Z 2 π r 2 d φ Ã Z φ 2 φ 1 sin φ d φ = 2 π r 2 Ã − cos φ ¯ ¯ ¯ ¯ cos − 1 a r φ = cos − 1 b r = 2 π r 2 µ − a r + b r ¶ = 2 π 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 π 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 θ , a sin θ , b ), ≤ θ ≤ 2 π 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 , θ ) = ( y cos θ , y sin θ , g ( y )) 12 for c ≤ y ≤ d , 0 ≤ θ ≤ 2 π . (c) Show that Area ( S ) = 2 π 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 θ , a sin θ , b ), ≤ θ ≤ 2 π (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 θ , y sin θ , g ( y )) , ≤ θ ≤ 2 π Therefore, the following parametrization parametrizes the surface S : 8( y , θ ) = ( y cos θ , y sin θ , g ( y )) , ≤ θ ≤ 2 π , 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 θ , y sin θ , g ( y )) =- cos θ , sin θ , g ( y ) ® T θ = ∂8 ∂ θ = ∂ ∂ θ ( y cos θ , y sin θ , g ( y )) = h− y sin θ , y cos θ , i The normal vector is their cross product:...
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17.4 Ex 43-51 - 1134 C H A P T E R 17 LINE AND SURFACE...

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