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Double integral bounds after polar transformation

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data

    Basically I'm just trying to convert a double integral into polar coordinates, but when I do it I get confused with my bounds.

    2. Relevant equations

    3. The attempt at a solution


    (x and y are just numbers, not variables). Then I transform the integral given polar coordinates [tex]u=r\cos\theta, v=r\sin\theta[/tex] and I get

    [tex]4\int_a^b\int_c^d e^{-r^2}(r\cos\theta)^{2x-1}(r\sin\theta)^{2y-1}rdrd\theta=4\int_a^b\int_c^d e^{-r^2}r^{2(x+y)-1}\cos^{2x-1}\theta\sin^{2y-1}\theta drd\theta[/tex].

    I know that [tex]a=0,b=\infty,c=0,d=\frac{\pi}{2}[/tex] so that the integral bounds are [tex]\int_0^{\infty}\int_0^{\pi/2}f(r,\theta)rdrd\theta[/tex], but I can't seem to understand the reasoning behind the bounds. Thanks for your help!
    Last edited: Dec 9, 2008
  2. jcsd
  3. Dec 9, 2008 #2
    In the uv-plane, the integral is over the first quadrant. Graph this region and describe it in terms of polar coordinates:

    Since the region extends infinitely, the radius r goes from 0 to infinity. Since the angle [tex]\theta[/tex] is 0 on the positive u-axis and [tex]\pi/2[/tex] on the positive v-axis, [tex]\theta[/tex] goes from 0 to [tex]\pi/2[/tex].

    So your bounds don't match up with their respective variables.
  4. Dec 9, 2008 #3


    Staff: Mentor

  5. Dec 9, 2008 #4
    Hrm, okay thank you. Are you saying that it's supposed to be [tex]\int_0^{\pi/2}\int_0^{\infty}f(r,\theta)rdrd\theta[/tex] rather than what I've put? Because I'm splitting this up into the product of integrals rather than evaluating the double integral for itself, so I get

    [tex]\int_0^{\infty}e^{-r^2}r^{2(x+y)-1}dr\cdot\int_0^{\pi/2}\cos^{2x-1}\theta\sin^{2y-1}\theta d\theta[/tex] in order to prove that [tex]\Gamma(x)\Gamma(y)=\Gamma(x+y)B(x,y)[/tex].
  6. Dec 10, 2008 #5


    Staff: Mentor

    Yes, because that's from the order in dr dtheta.
    If you want to integrate in the reverse order, I think that will work, too, because of the uncoupled nature of your f(r, theta), which can be written as g(r) * h(theta). That will, of course, switch the limits of integration.
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