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Math Help - Need help with double integral & polar coords

  1. #1
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    Need help with double integral & polar coords

    I've been stuck on this for a good while now, and it's really starting to irritate me. Any help I can get would be greatly appreciated.



    I've been trying to evaluate this problem for a good while now. I've looked through countless websites which explain the process of converting everything into polar coordinates, but for some reason it's just not clicking.

    The part I'm having the most problems with are the limits of integration. I just can't figure out how they're determined, and once I think I do have it figured out, things don't seem to work.

    Any help I can get is greatly appreciated.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Protoman View Post
    I've been stuck on this for a good while now, and it's really starting to irritate me. Any help I can get would be greatly appreciated.



    I've been trying to evaluate this problem for a good while now. I've looked through countless websites which explain the process of converting everything into polar coordinates, but for some reason it's just not clicking.

    The part I'm having the most problems with are the limits of integration. I just can't figure out how they're determined, and once I think I do have it figured out, things don't seem to work.

    Any help I can get is greatly appreciated.
    x = r~cos(\theta)
    and
    y = r~sin(\theta)

    So the integrand becomes
    x^2y^2 = r^4~sin^2(\theta)~cos^2(\theta)

    The area element becomes
    dx~dy = r~dr~d \theta

    And the region of integration is the right half-circle of radius 2 centered on the origin. So the integral becomes:
    \int_0^2 \int_{-\pi}^{\pi}r^4~sin^2(\theta)~cos^2(\theta) \cdot r~d \theta~dr

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post
    And the region of integration is the right half-circle of radius 2 centered on the origin.

    Thanks for the help...could you explain how you know this?
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  4. #4
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    Quote Originally Posted by Protoman View Post
    Thanks for the help...could you explain how you know this?
    Have you tried drawing the region of integration? x = \pm \sqrt{4 - x^2} \Rightarrow x^2 = 4 - y^2 \Rightarrow x^2 + y^2 = 2^2. So for 0 \leq y \leq 2 you clearly have a semi-circle centred on the origin. This clearly suggests using polar coordinates .....
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