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Math Help - Help with double integrals in polar coordinates

  1. #1
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    Help with double integrals in polar coordinates

    I appreciate any help on the following problems:

    1.) Find the area by double integration in polar coordinates inside circles r = 1 and r = 2sinΘ

    2.) Use double integration in polar coordinates to find the volume of solid that lies below given surface and above plane region R bounded by given curve: z=x^2 +y^2 and r=2cosΘ

    3.) Evaluate the given integral by first converting to polar coordinates: 1/(1+x^2 +y^2) dxdy where x is defined over region between 0 and (1-y^2)^1/2 and y is defined over region between 0 and 1.
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  2. #2
    Eater of Worlds
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    You could set the first one up like so.

    2\left[\int_{0}^{\frac{\pi}{6}}\int_{0}^{2sin{\theta}}rdr  d{\theta}+\int_{\frac{\pi}{6}}^{\frac{\pi}{2}}\int  _{0}^{1}rdrd{\theta}\right]

    Just one way to set it up.
    Last edited by galactus; November 21st 2009 at 12:57 PM.
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  3. #3
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    can u explain how u did that
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  4. #4
    Eater of Worlds
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    The first integral is the portion inside the circle r=2sin(t) and the second

    integral is inside the circle r=1. It may help if you look at a graph. You

    can see they intersect at Pi/6. The sum of the two give the area inside

    the two circles. Of course, double that result because that is half.
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