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Homework Help: Need help finding inverse transformation for Jacobian

  1. Mar 4, 2010 #1
    1. The problem statement, all variables and given/known data
    Suppose R is a plane region bounded by xy=1, xy=3, x^2-y^2=1, x^2-y^2= 4. Use the substitution u=xy, v=x^2-y^2 to evaluate

    [tex]I = \iint\limts_R \, (x^2+y^2) dx\,dy[/tex]

    3. The attempt at a solution
    Using the substitutions given, I find R` = { (u,v) | u for all [1,3] ; v for all [1,4] }

    So after this, it would seem I need to find the inverse transformation so I can make an appropriate substitution for the integrand and evaluate the jacobian so I can integrate over R`. However, I am having difficulties finding an explicit equation for both x and y in terms u and v. Here's what I've managed to do so far although its ultimately nothing... :(

    Squaring both sides of the equations relating u and v to x, y, I find:

    (I) [tex]u^2 = x^2y^2[/tex]
    (II) [tex]v^2 = x^4 - 2x^2y^2+y^4[/tex]

    Adding equations (I) + (II) I find:

    (III) [tex] v^2 + 2u^2 = x^4 +y^4[/tex]

    Dividing (II)/(I) equations I find:

    (IV) [tex] \frac{v^2}{u^2}\right) +2 = \frac{x^4+y^4}{x^2y^2}[/tex]

    And dividing equations (IV)/(III) I ultimately find:

    [tex]x^4+y^4=2(\frac{u^2v^2+2u^4}{v^2+2u^2}\right) )+ v^2[/tex]

    So as I said above, I really managed to find nothing at all. I've been fiddling around with both equations for a while now in hopes of finding a desirable "form" to be able to explicitly define both x and y in terms of u and v, but I am totally stumped... Any advice or ideas would be much appreciated. :)

  2. jcsd
  3. Mar 4, 2010 #2


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    What's v^2+4*u^2? You don't have to express x and y in terms of u and v. You just have to express x^2+y^2 in terms of u and v.
    Last edited: Mar 4, 2010
  4. Mar 4, 2010 #3
    So then how should I calculate the jacobian? I thought you needed x(u,v), y(u,v) to do so.
  5. Mar 4, 2010 #4


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    You can get that jacobian by calculating the jacobian of u(x,y) and v(x,y) and inverting it. Can't you?
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