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Thread: Triple Integral in Spherical Coords.

  1. #1
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    Triple Integral in Spherical Coords.

    Hey,
    I'm having difficulty with a specific problem. It was given on my professor's previous exam and is meant to be worked by hand but I'm getting a ridiculous integral that I don't think is solvable by hand in 10 mins (which it should be).

    It seems my formatting was lost so I attached a picture :s
    The cutoff part says (x^2+y^2)^1/2
    ill re upload


    Sorry for the terrible picture quality.

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  2. #2
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    I'm having difficulty reading the scanned handwriting, so I'm not sure how much of this you have already done. I'd start by converting x and y to polar coordinates and writing r = \sqrt{x^2+y^2}. Then the limits for z are r\leqslant z\leqslant\sqrt{4-r^2}. That implies that r^2\leqslant 4-r^2, or r\leqslant\sqrt2.

    So if you do the z-integral, and convert x and y to polars, then you get

    \begin{aligned}\int\!\!\!\int\!\!\!\int_Ex^2y^2z\,  dzdxdy &= \int_0^{2\pi}\int_0^{\sqrt2}r^2\cos^2\theta*r^2\si  n^2\theta \Bigl[\tfrac12z^2\Bigr]_r^{\sqrt{4-r^2}}\,rdrd\theta\\ &= \int_0^{2\pi}\tfrac12\cos^2\theta\sin^2\theta\Bigl[\tfrac23r^6 - \tfrac14r^8\Bigr]_0^{\sqrt2} = \frac23\int_0^{2\pi}\cos^2\theta\sin^2\theta\,d\th  eta. \end{aligned}

    Now use the fact that \cos^2\theta\sin^2\theta = \tfrac14\sin^2(2\theta) = \tfrac18(1-\cos(4\theta)) to do the theta-integral. I get the final answer as \frac{\pi}6.
    Last edited by Opalg; Apr 22nd 2009 at 07:21 AM.
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