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Math Help - area of the region that lies inside both curves

  1. #1
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    Sep 2007
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    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.

    <br />
  r_{1}=\sin\theta, r_{2}=\cos\theta<br />

    Which are the circles x^2+(y+\frac{1}{2})^2=\frac{1}{4} and (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:

    <br />
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}<br />
    <br />
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}<br />
    <br />
 A=A_{1}+A_{2}=(\frac{\pi}{16}-\frac{1}{8})+(\frac{\pi}{16}+\frac{1}{8})=\frac{\p  i}{8}<br />

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

    <br />
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}<br />

    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
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    Paris, France
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    I think for A_2 you would need to integrate from \frac{\pi}{4} to \frac{\pi}{2}.

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