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

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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. = 2 πf ( x ) ds (1) = 2 πf ( x ) 1 + ( f ( 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 ( x ) = - b a 2 x 1 1 - ( x a ) 2 . Plug into the formula: S.A. = 2 πy ( x ) 1 + ( y ( x )) 2 dx (3) = a - a 2 πb 1 - ( x a ) 2 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. First, I’m going to use a trick here that will really simplify things at the end. Notice that what we are integrating: 2 πb 1 - ( x a ) 2 1 + b 2 a 4 x 2 (1 - ( x a ) 2 ) - 1 is an even function. i.e. f ( - x ) = f ( x ). This means that it’s symmetric about the y -axis (also obvious from the picture cause it’s an ellipse). This means that
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