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Math Help - Polar Double Integral

  1. #1
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    Polar Double Integral

    Find the double integral of y(e^x) over the region in the first quadrant enclosed by the circle x^2+y^2=25.

    For this problem, I think I set up the integral properly, with the radius going from 0 to 5, and the angle going from 0 to pi/2. I set y=r*sin(theta) and x=r*cos(theta), but the first antiderivative from tabular integration by parts (w.r.t. r), was very messy, and the second antiderivative w.r.t. theta looks to be messy as well. I'm not sure whether to keep on brute forcing it, or whether there is a more elegant solution.

    I would appreciate any help. Mathematica returns 4e^5 - 23/2 as the answer by the way.
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  2. #2
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    Quote Originally Posted by MathTooHard View Post
    Find the double integral of y(e^x) over the region in the first quadrant enclosed by the circle x^2+y^2=25.

    For this problem, I think I set up the integral properly, with the radius going from 0 to 5, and the angle going from 0 to pi/2. I set y=r*sin(theta) and x=r*cos(theta), but the first antiderivative from tabular integration by parts (w.r.t. r), was very messy, and the second antiderivative w.r.t. theta looks to be messy as well. I'm not sure whether to keep on brute forcing it, or whether there is a more elegant solution.

    I would appreciate any help. Mathematica returns 4e^5 - 23/2 as the answer by the way.
    The integral will be, of course, \int \int (r \sin (\theta)) e^{r cos(\theta)} \, (r dr \, d\theta)= \int \int r^2 e^{r \cos(\theta)} \sin(\theta) \, d\theta \, dr. Integrate with respect to \theta first, by letting u= r \cos(\theta).
    Last edited by mr fantastic; November 10th 2009 at 10:43 PM. Reason: Fixed the latex
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  3. #3
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    Are you saying that I just have to deal with extremely messy antiderivatives?
    Last edited by mr fantastic; November 10th 2009 at 10:44 PM.
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  4. #4
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    Quote Originally Posted by MathTooHard View Post
    Are you saying that I just have to deal with extremely messy antiderivatives?
    I don't see the difficulty with integrating first with respect to \theta. You make a straightforward u-substitution. Where is your trouble here? (Note that HoI did not include the integral terminals, but they should be obvious ....)
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  5. #5
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    Oh thanks! The code didn't show up the first time so I didn't get it. I never thought of changing the orders of integration.
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