15_5_18 - < a < c , let k 2 = c 2-a 2 a 2 c 2 . Then...

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18. The upper half of the spheroid x 2 a 2 + y 2 a 2 + z 2 c 2 = 1 has a circular disk of radius a as projection onto the xy -plane. Since 2 x a 2 + 2 z c 2 z x = 0 z x = - c 2 x a 2 z , and, similarly, z y = - c 2 y a 2 z , the area element on the spheroid is dS = s 1 + c 4 a 4 x 2 + y 2 z 2 dx dy = s 1 + c 2 a 2 x 2 + y 2 a 2 - x 2 - y 2 dx dy = s a 4 + ( c 2 - a 2 ) r 2 a 2 ( a 2 - r 2 ) r dr d θ in polar coordinates. Thus the area of the spheroid is S = 2 a Z 2 π 0 d θ Z a 0 s a 4 + ( c 2 - a 2 ) r 2 a 2 - r 2 r dr Let u 2 = a 2 - r 2 u du = - r dr = 4 π a Z a 0 p a 4 + ( c 2 - a 2 )( a 2 - u 2 ) du = 4 π a Z a 0 p a 2 c 2 - ( c 2 - a 2 ) u 2 du = 4 π c Z a 0 s 1 - c 2 - a 2 a 2 c 2 u 2 du . For the case of a prolate spheroid 0
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Unformatted text preview: < a < c , let k 2 = c 2-a 2 a 2 c 2 . Then S = 4 π c Z a p 1-k 2 u 2 du Let ku = sin v k du = cos v d v = 4 π c k Z sin-1 ( ka ) cos 2 v d v = 2 π c k (v + sin v cos v) ± ± ± ± sin-1 ( ka ) = 2 π ac 2 √ c 2-a 2 sin-1 √ c 2-a 2 c + 2 π a 2 sq. units....
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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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