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Math Help - [SOLVED] Area computations in polar coordinates

  1. #1
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    [SOLVED] Area computations in polar coordinates

    Find the area of the region that lies interior to all three circles:
    r = 1
    r = 2cos \theta
    r = 2sin \theta

    I graphed it, but I have problems figuring out how to solve it. Any help with some explanations will be appreciated!
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  2. #2
    Junior Member roy_zhang's Avatar
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    Please see the attached graph. We have 3 circles: r=1 (in red), r=2\cos\theta (in blue) and r=2\sin\theta (in green). I labeled 2 intersection points of the common interior area of the three circles. They are (\frac{\sqrt{3}}{2},\frac{1}{2}) and (\frac{1}{2},\frac{\sqrt{3}}{2}) or in polar coordinates (1,\frac{\pi}{6}) and (1,\frac{\pi}{3})

    Look closely to the common interior region among the three circles, we know its area can be obtained by adding three areas together: 1) the area obtained by integrating the green circle ( r=2\sin\theta) from 0 to \frac{\pi}{6}, 2) the area obtained by integrating the red circle ( r=1) from \frac{\pi}{6} to \frac{\pi}{3}, 3) the area obtained by integrating the blue circle ( r=2\cos\theta) from \frac{\pi}{3} to \frac{\pi}{2}.

    Mathematically, we have:
    A=\frac{1}{2}\int\limits_0^\frac{\pi}{6} (2\sin\theta)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{6}^\frac{\pi}{3} (1)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{3}^\frac{\pi}{2} (2\cos\theta)^2 d\theta
    =\frac{5}{12}\pi-\frac{\sqrt{3}}{2}

    Note that the first and third area are identical, so we only need to calculate one of those.

    Roy
    Attached Thumbnails Attached Thumbnails [SOLVED] Area computations in polar coordinates-3_circles.png  
    Last edited by roy_zhang; March 21st 2008 at 11:56 AM.
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  3. #3
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    Thank you a lot! I was trying to split it into two parts rather than to three parts. Just one correction there in the integrals for people interested in the same problem:

    <br />
A=\frac{1}{2}\int\limits_0^\frac{\pi}{6} (2\sin\theta)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{6}^\frac{\pi}{3} (1)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{3}^\frac{\pi}{2} (2\cos\theta)^2 d\theta<br />
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  4. #4
    Junior Member roy_zhang's Avatar
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    Quote Originally Posted by hasanbalkan View Post
    Thank you a lot! I was trying to split it into two parts rather than to three parts. Just one correction there in the integrals for people interested in the same problem:

    <br />
A=\frac{1}{2}\int\limits_0^\frac{\pi}{6} (2\sin\theta)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{6}^\frac{\pi}{3} (1)^2 d\theta + \frac{1}{2}\int\limits_\frac{\pi}{3}^\frac{\pi}{2} (2\cos\theta)^2 d\theta<br />
    Hi hasanbalkan,

    Thank you for your correction! I corrected it.

    Roy
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