15_5_4 - θ r dr √ 4 a 2-r 2 Let u = 4 a 2-r 2 du =-2 r...

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4. One-quarter of the required area is shown in the figure. It lies above the semicircular disk R bounded by x 2 + y 2 = 2 ay , or, in terms of polar coordinates, r = 2 a sin θ . On the sphere x 2 + y 2 + z 2 = 4 a 2 , we have 2 z z x = - 2 x , or z x = - x z . Similarly, z y = - y z , so the surface area element on the sphere can be written dS = s 1 + x 2 + y 2 z 2 dx dy = 2 a dx dy p 4 a 2 - x 2 - y 2 . The required area is S = 4 ZZ R 2 a p 4 a 2 - x 2 - y 2 dx dy = 8 a Z π/ 2 0 d θ Z 2 a sin
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Unformatted text preview: θ r dr √ 4 a 2-r 2 Let u = 4 a 2-r 2 du = -2 r dr = 4 a Z π/ 2 d θ Z 4 a 2 4 a 2 cos 2 θ u-1 / 2 du = 8 a Z π/ 2 ( 2 a-2 a cos θ) d θ = 16 a 2 (θ-sin θ) ± ± ± π/ 2 = 8 a 2 (π-2 ) sq. units. x y z z 2 = 4 a 2-x 2-y 2 2 a 2 a r = 2 a sin θ 2 a Fig. 5-4...
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This note was uploaded on 03/08/2012 for the course MATH 120 taught by Professor Onurfen during the Spring '12 term at Middle East Technical University.

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