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Math Help - Convert an integral from Cartesian to Polar coordinates and evaluate it.

  1. #1
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    Convert an integral from Cartesian to Polar coordinates and evaluate it.



    Need help with this.
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by asdf122345 View Post


    Need help with this.
    Convert an integral from Cartesian to Polar coordinates and evaluate it.-graph223.jpg

    r^2 = x^2 +y^2
    y = r\sin \theta
    x = r \cos \theta
    lets convert the boundaries

    x = \sqrt{4y-y^2 }

    x^2 = 4y -y^2 \Rightarrow x^2+y^2 = 4y

    r^2 = 4r\sin \theta

    r = 4 \sin \theta

    r changes from 0 to 4\sin \theta

    and theta from 0 to pi/2 the first quarter so

    \int_{0}^{4}\int_{0}^{\sqrt{4y-y^2}} x^2 \; dx dy

    becomes

    \int_{0}^{\frac{\pi}{2}} \int_{0}^{4\sin \theta} r^2 \cos^2 \theta  (r) dr d\theta

    \int_{0}^{\frac{\pi}{2}} \cos^2 \theta \left(\frac{r^4}{4}\right)\mid^{4 \sin \theta}_{0} d\theta

    \int_{0}^{\frac{\pi}{2}} \cos^2\theta (4^3 \sin^4 \theta ) d\theta
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  3. #3
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    Is it possible to integrate where you left off? I can't figure out how to integrate that. Or does it get messy from integrating from that.
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  4. #4
    MHF Contributor Amer's Avatar
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    4^3 \int \cos ^2 \theta (\sin ^4 \theta ) d\theta

    \sin  \theta \cos  \theta = \frac{\sin 2\theta }{2}

    \sin ^2 \theta = \frac{1-\cos 2\theta}{2}

    4^3 \int \cos ^2 \theta \sin ^2 \theta ( \sin ^2\theta ) d \theta


    4^3 \int \frac{\sin ^2 2\theta}{4} \left(\frac{1- \cos 2\theta }{2}\right) d\theta

    4^3 \left(\int \frac{\sin ^2 2\theta}{4} d\theta - \int \frac{\sin ^2 2 \theta \cos 2\theta}{8} d\theta\right)

    first integral convert \sin ^2 2\theta = \frac{1- \cos 4 \theta }{2}

    second one sub u = \sin 2\theta

    can you continue ?
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  5. #5
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    Yea I got it from here. Thanks
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  6. #6
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    I got 4 pi as the answer. Is that correct? Something seems wrong.
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  7. #7
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    The answer is 2\pi

    So you may have just made a small error by a factor of 2.
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