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Math Help - Help in setting the polar integral

  1. #1
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    Help in setting the polar integral

    Q: Find the area inside the region bounded by both r = sin theta and r = sin (2 theta)

    I set the integral to.........

    2[1/2{(sin 2x)^2 dx - 1/2{(sin2x)^2 - (sinx)^2 dx]

    Notes: For efficiency, I used x for theta. Also, the limits of integration for each integral are from 0 to pi/3.

    But my book says this, which doesn't make sense -

    2[1/2{(sinx)^2 dx + 1/2{(sin2x)^2 dx]

    Limits of integration for first integral: from 0 to pi/3
    Limits of integration for second integral: from pi/3 to pi/2

    Help please!

    Cheers!
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  2. #2
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    Believe it or not, your book is right. The area swept out by a radius r through angle \theta is given by \frac{1}{2}r^2\theta so your differential of area is \frac{1}{2}r^2 d\theta

    For \theta from 0 to \pi/3, sin(2\theta)> sin(\theta) so the radius only extends from 0 to sin(\theta). For \theta from \pi/3 to \pi/2 sin(2\theta) is less than sin(\theta) so the radius extends only to that.
    The area between the two curves in the first quadrant is \frac{1}{2}\int_0^{\pi/3} sin^2(\theta)d\theta+ \int_{\pi/3}^{\pi/2} sin^2(2\theta)d\theta.

    Of course, there is a symmetrical part in the second quadrant so that is multiplied by 2.

    What you appear to be calculating is the area outside r= sin(\theta) and inside r= sin(2\theta).
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  3. #3
    MHF Contributor
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    Hi

    Have a look to the sketch below


    Let's focus on the first quadrant
    First you need to find where the 2 curves cross
    \sin(2\theta) = \sin(\theta)
    \sin(\theta) (2 \cos(\theta) - 1) = 0
    \theta = 0 or \theta = \frac{\pi}{3}

    Then you need to sweep the area
    You can see that from \theta = 0 to \theta = \frac{\pi}{3}, r goes from 0 to \sin(\theta) (green lines)
    And from \theta = \frac{\pi}{3} to \theta = \frac{\pi}{2}, r goes from 0 to \sin(2\theta) (blue lines)
    Last edited by running-gag; February 13th 2009 at 11:29 AM. Reason: Correction of error
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  4. #4
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    Unbelievable, you guys are the best! I'm still somewhat inexperienced with the polar form.
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