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Math Help - Area polar coordinates.

  1. #1
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    Area polar coordinates.

    Hi. I wanted to know if I did this right:
    Using double integrals, set in polar coordinates, the integral that gives the area for the region D: \{x^2+y^2\leq{9},(x-2)^2+y^2\geq{7}

    From the picture:


    The pink region represents the area I want.
    The intersection is given for x=3/2, where: y=\displaystyle\frac{3\sqrt[ ]{3}}{2}

    Then \tan \theta_1=9\sqrt[ ]{3}

    Tita 1 represents the limit on the angle for the circle that isn't centered at the origin. Is the limit for the brown region.

    \tan \theta_2=3\displaystyle\frac{\sqrt[ ]{3}}{4}
    This is the limit on the angle for the orange region, for the circle centered at the origin.

    Finally
    y=\displaystyle\frac{-3\sqrt[ ]{3}}{2}(x+3)
    With x=rcos\theta,y=r\sin \theta
    Then r=\frac{-15\sqrt[ ]{3}}{2(\sin\theta+\frac{3\sqrt[ ]{3}}{2}\cos\theta)}

    And the area:

    \displaystyle\int_{0}^{2\pi}\int_{0}^{3}rdrd\theta-2\displaystyle\int_{0}^{tan^{-1}9\sqrt[ ]{3}}\int_{0}^{\sqrt[ ]{7}}rdrd\theta-2\displaystyle\int_{0}^{tan^{-1}\displaystyle\frac{3\sqrt[ ]{3}}{4}}\int_{\frac{-15\sqrt[ ]{3}}{2(\sin\theta+\frac{3\sqrt[ ]{3}}{2}\cos\theta)}}^{3}rdrd\theta

    The first integral represents the entire circle, the second the brown region and the third the orange region, both multiplied on two by simmetry
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  2. #2
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    Re: Area polar coordinates.

    My suggestion is to integrate by having r go from the brown circle to the blue circle as θ goes from θ_1 to π, then double the answer.
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  3. #3
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    Re: Area polar coordinates.

    How would you do that? I tried but I couldn't get an expression for r.
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  4. #4
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    Re: Area polar coordinates.

    In polar coordinates, the equation of the larger circle is obvious, hopefully.

    Find the equation of the smaller circle in polar coords.

    (x-2)^2+y^2=7

    x^2-4x+4+y^2=7

    r^2-4r\cos(\theta)-3=0

    Solve for r using the quadratic formula or by completing the square.

    We usually like to have r be positive, so use the positive sign with the square root. (Either sign will give the same polar graph.)
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  5. #5
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    Re: Area polar coordinates.

    Great, didn't realized about that. That simplifies things pretty much. Anyway, the answer I give is right or did I made a mistake? Just wanna know, I know its horrendous, but anyway I'd like to know if instead of that its ok.

    This is the solution I get using your suggestion:

    \displaystyle\int_0^{2\pi}\int_{\frac{-2\cos\theta+\sqrt[]{4\cos^2\theta+12}}{2}}^{3} rdrd\theta
    Nice
    Last edited by Ulysses; August 17th 2011 at 04:58 AM.
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  6. #6
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    Re: Area polar coordinates.

    You have some algebra error(s) in your solution to the quadratic equation.
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  7. #7
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    Re: Area polar coordinates.

    You're right, I've made some mistakes. The lower bound would be \frac{4\cos\theta+\sqrt[]{16\cos^2\theta+12}}{2}

    Thank you very much SammyS.
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