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Math Help - Double Integral In Polar Coordinates

  1. #1
    Junior Member
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    Double Integral In Polar Coordinates

    Hi all,

    I'm trying to solve the following by converting it to polar coordinates.

    \int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}} xy dxdy

    I know that the x*y term will become:

    (rcos(\theta))(rsin(\theta))r drd\theta

    However I'm not sure how to get the integration boundaries. I've looked around but I can't find an example that is similar. Any help would be greatly appreciated. Thanks.
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  2. #2
    MHF Contributor
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    What about this integral,

    \displaystyle{\int_{0}^{\sqrt{2}}\int_{y}^{\sqrt{4-y^2}}\ 1\ dx\ dy}

    Can you draw the area? If not, notice that the above is another way of writing

    \displaystyle{\int_{0}^{\sqrt{2}}\sqrt{4-y^2}}\ -\ y\ dy}

    So it's between the lines/curves for \displaystyle{x = \sqrt{4-y^2}} and x = y and y = 0 and y = \sqrt{2}. So it's a nice sector of a circle, which gives you the polar limits.
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  3. #3
    Senior Member yeKciM's Avatar
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    which looks like :
    Attached Thumbnails Attached Thumbnails Double Integral In Polar Coordinates-doubleintegral.jpg  
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