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Math Help - Using polar coordinates to solve a double integral

  1. #1
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    Using polar coordinates to solve a double integral

    Hi,
    I am confused on how to go from rectangular components to polar ones. I'm going to show you the problem I'm trying to solve and the steps I took to solve it. I would really be glad if anyone could take a look and tell me if I'm doing something wrong.
    Thanks for viewing,
    Bazinga

    The question is:
    Use polar coordinates to solve the following integral.

    \int^{3}_{-3} \int^{\sqrt{9-y^2}}_{-\sqrt{9-y^2}} e^{-x^2-y^2} \mathrm{d}x\mathrm{d}y

    I thought that \sqrt{9-y^2} and -\sqrt{9-y^2} formed two semi-circles (the positive one in the 1st and 4th quadrant, the other one in the 2nd and 3rd quadrant), which formed a circle.

    Then the double integral in polar coordinates would be:

    \int^{2\pi}_{0} \int^{3}_{0} e^{-r^2}r \mathrm{d}r\mathrm{d}\theta

    Is that right?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Right.
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  3. #3
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    Thank you very much! I'll mark this post as solved as it could help somebody else in the future.
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