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Thread: area of the region that lies inside both curves

  1. #1
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    Sep 2007
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    6

    area of the region that lies inside both curves

    Hi,

    I am trying to solve this problem:

    Find the area of the region that lies inside both curves.

    $\displaystyle
    r_{1}=\sin\theta, r_{2}=\cos\theta
    $

    Which are the circles $\displaystyle x^2+(y+\frac{1}{2})^2=\frac{1}{4}$ and $\displaystyle (x+\frac{1}{2})^2+y^2=\frac{1}{4}$ on the Cartesian coordinate system.

    When I try to solve the problem in the following way, I get a different answer than in my textbook:

    $\displaystyle
    A_{1}=\int_0^\frac{\pi}{4}\frac{1}{2}\sin^2\theta\ d\theta=\frac{1}{4}\int_0^\frac{\pi}{4}(1-\cos2\theta)\ d\theta=\frac{1}{4}(\frac{\pi}{4}-\frac{1}{2})=\frac{\pi}{16}-\frac{1}{8}
    $
    $\displaystyle
    A_{2}=\int_0^\frac{\pi}{4}\frac{1}{2}\cos^2\theta\ d\theta=\frac{1}{4}\int_0^\frac{\pi}{4}(1+\cos2\th eta)\ d\theta=\frac{1}{4}(\frac{\pi}{4}+\frac{1}{2})=\fr ac{\pi}{16}+\frac{1}{8}
    $
    $\displaystyle
    A=A_{1}+A_{2}=(\frac{\pi}{16}-\frac{1}{8})+(\frac{\pi}{16}+\frac{1}{8})=\frac{\p i}{8}
    $

    However, if I do it by symmetry, it gives me a different answer (the correct answer in my textbook):

    $\displaystyle
    A=2\int_0^\frac{\pi}{4}\frac{1}{2}\sin^2\theta\ d\theta=\frac{1}{2}\int_0^\frac{\pi}{4}(1-\cos2\theta)\ d\theta=\frac{1}{2}(\frac{\pi}{4}-\frac{1}{4})=\frac{\pi}{8}-\frac{1}{4}
    $

    Can anyone explain this to me? I've been trying to figure it out, but I can't seem to get it.
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  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    I think for $\displaystyle A_2$ you would need to integrate from $\displaystyle \frac{\pi}{4}$ to $\displaystyle \frac{\pi}{2}$.

    Laurent.
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  3. #3
    Newbie
    Joined
    Sep 2007
    Posts
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    I see it now. Thanks. =)
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