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Math Help - integral.

  1. #1
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    integral.

    Does anyone know how to do this integral?
    \int_{x=0}^{x=3}\int_{y=0}^{y=\sqrt{9-x^2}}\sqrt{1+4x^2+4y^2}\,dy\,dx I think a substituition you can use is either sin or cos but i'm not too sure.
    Last edited by oxrigby; November 18th 2009 at 01:37 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oxrigby View Post
    Does anyone know how to do this integral?
    \int_{x=0}^{x=3}\int_{y=0}^{y=\sqrt{9-x^2}}\sqrt{1+4x^2+4y^2}\,dy\,dx I think a substituition you can use is either sin or cos but i'm not too sure.
    Rewrite this as \int_0^3\int_0^{\sqrt{9-x^2}}\sqrt{1+4\left(x^2+y^2\right)}\text{ }dy\text{ }dx. Make a change of coordinates to polar.(don't forget the Jacobian) so that we get \int_0^{\frac{\pi}{2}}\int_0^3 r\cdot\sqrt{1+4r^2}\text{ }dr\text{ }d\theta
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  3. #3
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by oxrigby View Post
    Does anyone know how to do this integral?
    \int_{x=0}^{x=3}\int_{y=0}^{y=\sqrt{9-x^2}}\sqrt{1+4x^2+4y^2}\,dy\,dx I think a substituition you can use is either sin or cos but i'm not too sure.
    Use polar coordinates!!!

    x=r\cos\theta
    y=r\sin\theta
    dA=r\,dr\,d\theta

    Here your area is the quarter circle in the first quadrant, so the bounds on \theta are 0..\frac{\pi}{2}.

    Your circle has radius 3, so the bounds on r are 0..3.

    So you're integrating

    \int_0^{\pi/2}\int_0^3 r\sqrt{1+4r^2}\,dr\,d\theta

    which is straightforward.
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