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Thread: 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 $\displaystyle \theta$ is given by $\displaystyle \frac{1}{2}r^2\theta$ so your differential of area is $\displaystyle \frac{1}{2}r^2 d\theta$

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

    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 $\displaystyle r= sin(\theta)$ and inside $\displaystyle r= sin(2\theta)$.
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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
    $\displaystyle \sin(2\theta) = \sin(\theta)$
    $\displaystyle \sin(\theta) (2 \cos(\theta) - 1) = 0$
    $\displaystyle \theta = 0$ or $\displaystyle \theta = \frac{\pi}{3}$

    Then you need to sweep the area
    You can see that from $\displaystyle \theta = 0$ to $\displaystyle \theta = \frac{\pi}{3}$, r goes from 0 to $\displaystyle \sin(\theta)$ (green lines)
    And from $\displaystyle \theta = \frac{\pi}{3}$ to $\displaystyle \theta = \frac{\pi}{2}$, r goes from 0 to $\displaystyle \sin(2\theta)$ (blue lines)
    Last edited by running-gag; Feb 13th 2009 at 10:29 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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