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Math Help - Messy Surface Integral

  1. #1
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    Question Messy Surface Integral

    Hello all I'm trying to do this surface integral and I'm getting really nasty numbers.

    This was taken from an old test and the results on these tests usually work out to a reasonable clean number compared to what I've gotten as an answer.

    Did I make any mistakes? Am I doing this the hard way? Anything wrong?

    Thanks again!
    Attached Thumbnails Attached Thumbnails Messy Surface Integral-2009q2at2.jpg  
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  2. #2
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    If you are trying to integrate over the cured surface then a change of variables will make this much easier.

    Choose x = 2*Cos phi, y=2*sin phi. I.e. change to cylindrical coordinates.

    Thenyou must integrate phi from 0 to pi/4 and z from 0 to 2.

    ie
    \int^2_0 \int^{pi/4}_0 4Cos(\phi)Sin(\phi)d \phi dz

    This integral is quite easy to solve.
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  3. #3
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    Quote Originally Posted by Kiwi_Dave View Post
    If you are trying to integrate over the cured surface then a change of variables will make this much easier.

    Choose x = 2*Cos phi, y=2*sin phi. I.e. change to cylindrical coordinates.

    Thenyou must integrate phi from 0 to pi/4 and z from 0 to 2.

    ie
    \int^2_0 \int^{pi/4}_0 4Cos(\phi)Sin(\phi)d \phi dz

    This integral is quite easy to solve.
    No, the boundary is the parabola [itex]y= 4- x^2[/itex], not the circle [itex]y^2= 4- x^2[/itex] as you seem to think.

    Jegues, your result is what I get.
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