math31b_hwprob14

math31b_hwprob14 - p595, number 29. Find the surface area...

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p595, number 29. Find the surface area of the ellipse x 2 a 2 + y 2 b 2 = 1 for a > b when rotated about x -axis. —————————————— Notice that this ellipse has the points ( - a, 0) and (+ a, 0) as its x -intercepts and ( - b, 0), (+ b, 0) as y -intercepts. Check this. So solve for y ( x ) and plug into the formula: S.A. = Z 2 πf ( x ) ds (1) = Z 2 πf ( x ) p 1 + ( f 0 ( x )) 2 dx (2) And then let x go from - a to a . Solve for y : y ( x ) = bsqrt 1 - ( x a ) 2 . Differentiate with respect to x y 0 ( x ) = - b a 2 x 1 p 1 - ( x a ) 2 . Plug into the formula: S.A. = Z 2 πy ( x ) p 1 + ( y 0 ( x )) 2 dx (3) = Z a - a 2 πb r 1 - ( x a ) 2 r 1 + b 2 a 4 x 2 (1 - ( x a ) 2 ) - 1 dx. (4) All we need to do now is compute the integral. It’s a pain. There will be lots of u -sub. and notice how the bounds on the integral keep changin’ each time. It’s okay to not keep track as long as we go back to x at the end.
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This note was uploaded on 07/11/2009 for the course MATH Math 31B taught by Professor Houdayer during the Winter '09 term at UCLA.

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math31b_hwprob14 - p595, number 29. Find the surface area...

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