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Change of variable integral problem

  1. Jun 7, 2012 #1
    1. The problem statement, all variables and given/known data
    http://img716.imageshack.us/img716/7453/28042782.jpg [Broken]

    3. The attempt at a solution

    What I do not understand is how the Jacobian suddenly gets inverted when doing the integral, I have looked over my other tutorial problems for similar solutions and I do not recall doing something like this.

    http://img140.imageshack.us/img140/7459/45340834.jpg [Broken]
    http://img220.imageshack.us/img220/7548/63185150.jpg [Broken]
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Jun 7, 2012 #2

    vela

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    The Jacobian calculated in the solution is
    $$J = \begin{vmatrix}
    \partial u/\partial x & \partial u/\partial y \\
    \partial v/\partial x & \partial v/\partial y
    \end{vmatrix}$$ so you have that ##du\,dv = J\,dx\,dy##. Because you want to find ##\int dx\,dy##, you need to integrate ##\int (1/J)\,du\,dv##. If you used the inverse transformations and found instead
    $$J' = \begin{vmatrix}
    \partial x/\partial u & \partial x/\partial v \\
    \partial y/\partial u & \partial y/\partial v
    \end{vmatrix},$$ then you'd have ##dx\,dy = J'\,du\,dv##.
     
  4. Jun 7, 2012 #3
    Yes, it looks like velas response is correct, but let me add a notation that might help you remember which way you are going. There is a visual believability about

    [tex]dudv=|\frac{\partial(u,v)}{\partial(x,y)}|dxdy[/tex]

    and so also for

    [tex]dxdy=|\frac{\partial(x,y)}{\partial(u,v)}|dudv[/tex]

    See? It sort of looks like things cancel correctly.
     
    Last edited: Jun 7, 2012
  5. Jun 7, 2012 #4

    SammyS

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    vela has a history of giving correct responses!
     
  6. Jun 8, 2012 #5
    Thanks for the responses, yea I sort of see it now. I guess a bit more practice with it is required.
     
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