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Jacobian and Multiple Integral

  1. Jul 27, 2009 #1
    I was self-studying the Jacobian and the change in variables when I came upon the following problem:

    In the integral

    [tex] I = \int_0^\infty \int_0^\infty \frac{x^2 + y^2}{1 + (x^2 - y^2)^2} e^{-2xy} \, dx dy , [/tex]

    make the change of variables

    [tex] u = x^2 - y^2 , [/tex]
    [tex] v = 2xy , [/tex]

    and evaluate I.

    I know that I have to use the formula

    [tex] I = \int_0^\infty \int_0^\infty |J| f(u, v) \, du dv [/tex]

    where |J| is the determinant of the Jacobian. But I don't know how to calculate the elements of the Jacobian. For example, how do I calculate [tex] {\partial x}/{\partial u} [/tex] if x is an implicit variable in the equation for u? Do I use implicit differentiation (if there is one) or do I take the positive square root of x^2 then take its partial derivative?

    Finally, how do I set up the bounds for the new integral? My way of thinking is if x = 0, then u = -y^2 and if x = infinity then u = infinity. I really feel there is something wrong with this. Also, some people would do it geometrically and graph to see the bounds. Is there any way to get the new bounds algebraically? Thank you very much! ^^

    (By the way, the answer at the back is [tex] \pi / 4 [/tex]). =P
  2. jcsd
  3. Jul 27, 2009 #2


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    Do the Jacobian the other way:
    [tex]\frac{\partial u,v}{\partial x,y}= \left|\begin{array}{cc}2x & -2y \\ 2y & 2x\end{array}\right|[/tex]
    [tex]= 4x^2+ 4y^2[/tex]
    so that
    [tex]|J|= \frac{\partial x,y}{\partial u,v}= \frac{1}{4}\frac{1}{x^2+ y^2}[/tex]

    Fortunately, you have [itex]x^2+ y^2[/itex] in your integral to cancel that!
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