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Change of variables jacobian limits

  1. Aug 17, 2015 #1
    ## \int_{0}^{∞}\int_{0}^{∞} \frac{x^2+y^2}{1+(x^2-y^2)^2} e^{-2xy} dxdy ##

    ##u= x^2-y^2##
    ##v=2xy##


    I tried to find the jacobian and the area elements,

    I found it to be ## dA = \frac{1}{v} du dv ##

    I'm having problem finding the limits of u & v and getting rid of ##x^{2}+y^{2}##.
     
  2. jcsd
  3. Aug 17, 2015 #2
    ------------- by mistake
     
  4. Aug 17, 2015 #3

    Zondrina

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    I could be wrong about this, but I think you should try a polar transformation. All the ##x^2## and ##y^2## terms seem to hint at this.

    $$\iint_D \frac{x^2 + y^2}{1 + (x^2 - y^2)^2} e^{-2xy} dA = \iint_{D'} \frac{r^2}{1 + (r^2 \text{cos}(2 \theta))^2} e^{-r^2 sin(2 \theta)} r dr d\theta$$

    Where ##D'## is the part of ##D## inside the circle ##x^2 + y^2 \leq c^2##. So ##0 \leq r \leq c## and ##0 \leq \theta \leq \frac{\pi}{2}## define the limits of ##D'##.

    Then you want to consider when ##c \to \infty##, so ##D'## becomes infinitely large.
     
    Last edited: Aug 17, 2015
  5. Aug 18, 2015 #4
    Hmmm, seems good. But the question require using Jacobian determinant.
     
  6. Aug 18, 2015 #5

    Zondrina

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    When changing to polar co-ordinates:

    $$dA = \left| J \right| dA' = \left| J \right| dr d\theta = r \space dr d\theta$$

    The Jacobian is just ##r##.
     
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