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Math Help - area; polar coordinates

  1. #1
    Member billym's Avatar
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    area; polar coordinates

    I have \int\int_Rx^3+xy^2dxdy,

    R = second quadrant of 1/4<=x^2+y^2<=1



    I changed x^3+xy^2 to polar coordinates:

    r^3cos^3(\theta)+r cos(\theta)r^2sin^2(\theta)=r^3cos(\theta)

    Do I evaluate the integral like this?

    \int_{\pi/2}^{\pi}\int_{1/2}^{1}r^3cos(\theta)drd\theta
    Last edited by billym; November 13th 2009 at 05:31 AM.
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  2. #2
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    I think your boundaries are correct.
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  3. #3
    Eater of Worlds
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    Don't forget about that extra r when using polar integration.

    x^{3}+xy^{2}

    r^{3}cos^{3}{\theta}+rcos{\theta} \cdot r^{2}sin^{2}{\theta}=r^{3}cos{\theta}

    r \cdot r^{3}cos{\theta}
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  4. #4
    Member billym's Avatar
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    Should the answer be the same as \frac{\pi-\pi \frac{1}{2}^2}{4} or am I missing the point here?
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  5. #5
    Member billym's Avatar
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    Or rather, that is R. Which is basically the upper left portion of a doughnut. But what does the area of integration look like?
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  6. #6
    Member billym's Avatar
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    Pardon me for being the village idiot, but could somebody please answer this one question for me:

    When sketching the area of integration, do I just sketch R , or do I have to consider x^3+xy^2 ?
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  7. #7
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    Quote Originally Posted by billym View Post
    Should the answer be the same as \frac{\pi-\pi \frac{1}{2}^2}{4} or am I missing the point here?
    The answer shouldn't have pi in it. \int_{\frac{1}{2}}^1 \int_{\frac{\pi}{2}}^\pi r^4 \cos (\theta ) d\theta dr = \int_{\frac{1}{2}}^1 -r^4 dr.
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  8. #8
    Member billym's Avatar
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    yeah I realized that last night. Anyway, I've got the right answer now. Thanks dudes.
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