15_5_19 - v) ) ± ± ± ± tan-1 ( ka ) = 2 π ac 2 √ a...

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19. We continue from the formula for the surface area of a spheroid developed part way through the solution above. For the case of an oblate spheroid 0 < c < a , let k 2 = a 2 - c 2 a 2 c 2 . Then S = 4 π c Z a 0 p 1 + k 2 u 2 du Let ku = tan v k du = sec 2 v d v = 4 π c k Z tan - 1 ( ka ) 0 sec 3 v d v = 2 π c k ( sec v tan v + ln ( sec v + tan
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Unformatted text preview: v) ) ± ± ± ± tan-1 ( ka ) = 2 π ac 2 √ a 2-c 2 " a √ a 2-c 2 c 2 + ln a c + √ a 2-c 2 c !# = 2 π a 2 + 2 π ac 2 √ a 2-c 2 ln a + √ a 2-c 2 c ! 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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