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Math Help - double integrals in polar coordinates

  1. #1
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    double integrals in polar coordinates



    And this is the answer they have given us:




    They converted to polar coordinates. I've seen this type of problem being solved without conversion to polar coordinates. I tried to evaluate the integral without converting to polar and I ended up with a different answer:

    \int^1_0 \int^{-y+1}_{x=0} 2x-2y dx dy

    \int^1_0 y^2-2y+1dy = \frac{1}{3}-1+1=\frac{1}{3}

    But my answer does not agree with the model answer! Shouldn't we end up with the same result regardless of whether we use polar or rectangular coordinates? I appreciate any help because I have an exam tomorrow...
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    Last edited by demode; June 21st 2010 at 10:43 PM.
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  2. #2
    Junior Member autumn's Avatar
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    You left out the r with the jacobian or the r associated with the x or y.
    In either case you need two r's in the integral

    x=r\cos\theta and y=r\sin\theta while dxdy=rdrd\theta

    Furthermore, you shoudn't go anywhere near polar, since this is a triangle and your polar bounds are of a circle.

    the line x+y=1 converted to polar is hideous.... r={1\over \sin\theta+\cos\theta}
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