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Math Help - Area of polar curve

  1. #1
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    Area of polar curve

    Find the area of the region that is inside r^2=2sin2\theta and inside r=1

    Attempt to solution:


    r^2=r^2

    2sin2\theta=1

    \theta= \frac{\pi}{12}, \frac{5\pi}{12}

    <br />
\int_\frac{\pi}{12}^\frac{5\pi}{12} \frac{1}{2}((2sin2\theta)^2-(1)^2)d\theta)<br />


    I don't understand how this is wrong. My answer is totally different from the one in the book
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  2. #2
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    Quote Originally Posted by nyasha View Post
    Find the area of the region that is inside r^2=2sin2\theta and inside r=1

    Attempt to solution:


    r^2=r^2

    2sin2\theta=1

    \theta= \frac{\pi}{12}, \frac{5\pi}{12}

    <br />
\int_\frac{\pi}{12}^\frac{5\pi}{12} \frac{1}{2}((2sin2\theta)^2-(1)^2)d\theta)<br />


    I don't understand how this is wrong. My answer is totally different from the one in the book
    note the symmetrical areas in quads I and III ...

    A = 4\int_0^{\frac{\pi}{12}} \sin(2\theta) \, d\theta + 4\int_{\frac{\pi}{12}}^{\frac{\pi}{4}} \frac{1}{2} \, d\theta
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