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Math Help - Polar Area using a Double Integral

  1. #1
    Member Em Yeu Anh's Avatar
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    Polar Area using a Double Integral

    Q: Obtain the area inside r=1+cos\theta and outside r=3cos\theta

    My first attempt was A=2\int_{\frac{\pi}{3}}^{\pi}\int_{3cos\theta}^{1+  cos\theta}rdrd\theta which I quickly realized has incorrect angles, since r=3cos\theta loops around twice as fast as the other graph. Thank you!

    EDIT - I have just figured it out on my own! Sorry about that
    Last edited by Em Yeu Anh; May 25th 2010 at 08:59 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    You need two integrals.

    The circle r=3\cos\theta doesn't reach the second quadrant.

    So you integrate from \pi/3 to \pi/2 with the bounds you used

    BUT in the second integral you integrate from \pi/2 to \pi

    where the lower bound is zero not 3\cos\theta
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