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Math Help - Double interal in polar coordinates

  1. #1
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    Double interal in polar coordinates

    I'm trying to evaluate \iint_D {(4-x^2-4y^2)}\, dA over the region enclosed by the ellipse D: \frac{x^2}{4} + y^2 = 1.

    Transforming the region to a circle of radius 1 (with the transformation u=\frac{x}{2}, v=y) and then to polar coordinates gives the result 4 \pi, which corresponds to the key of the problem.

    However, I figured I would attempt to do the integration again by going directly to polar coordinates, giving, after a bit more work

    \iint_D {(4-x^2-4y^2)}\, dA = \int_0^{2 \pi} d\theta \int_0^{\sqrt{4\cos^2{\theta}+\sin^2{\theta}}} {(4-r^2 \cos^2{\theta} - 4 r^2 \sin^2{\theta})r}\, dr = ... = \frac{115 \pi}{32}.

    I checked the latter result by integrating the integral after the first equality sign in the row above in Mathematica, giving \frac{115 \pi}{32}. Thus, it seems likely that there's some conceptual misunderstanding on my part. What exactly did I do wrong?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    The elipse x^2/4+y^2=1 in polar coordinates is \rho^2\cos^2 \theta+4\rho^2\sin^2 \theta=4 , so:


    0\leq \rho \leq \sqrt{\dfrac{4}{\cos ^2\theta+4\sin^2\theta}}
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  3. #3
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    Ah, indeed. I was too hasty, thinking of r being integrated from 0 to the length of the radius in Euclidean coordinates.

    Thank you!
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