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Change of variables Jacobian

  1. Dec 12, 2005 #1
    Hi, I have the following integral.

    \int\limits_{}^{} {\int\limits_R^{} {\left( {\sinh ^2 x + \cos ^2 y} \right)} \sinh 2x\sin 2ydxdy}

    Where R is the part of the region 0 <= x, 0 <= y <= pi/2 bounded by the curves x = 0, y = 0, sinhxcosy = 1 and coshxsiny = 1.

    In the hints section, there is a part which says [tex]J_{xy,uv} = \left( {\sinh ^2 x + \cos ^2 y} \right)^{ - 1} [/tex].

    Firstly, to evaluate this integral I need to make a change of variables. The obvious ones are u = sinhxcosy and v = coshxsiny. Usually, to compute the Jacobian I would find expressions for x and y in terms of u and v. In this case this doesn't look possible.

    The hint seems to have used [tex]\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}} = \left[ {\frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}}} \right]^{ - 1} [/tex]. I know this is valid for some cases but I'm not sure which ones. Can someone explain to me when I can use the Jacobian relation given above?

    Any help is appreciated.
  2. jcsd
  3. Dec 12, 2005 #2


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    As long as it isn't 0! The Jacobian is the determinant of the transformation matrix. The determinant of A-1 is one over the determinant of A. The only problem you might have is that you will want the integral written in terms of the new variable u and v.
  4. Dec 12, 2005 #3
    I don't really understand their notation with the Js but from my working, if u = sinhxcosy and v = coshxsiny then:

    \frac{{\partial \left( {u,v} \right)}}{{\partial \left( {x,y} \right)}} = \sinh ^2 x + \cos ^2 y

    If I take [tex]\frac{{\partial \left( {x,y} \right)}}{{\partial \left( {u,v} \right)}}[/tex] to be the reciprocal of the above then the integral is easy to evaluate.

    I just had my doubts because in most of the solutions I've seen, the author has avoided using the above. Rather, an effort is always made to solve equations to get the new variables in terms of the original variables so that the Jacobian is calculated directly. Anyway thanks for the help.
  5. Dec 16, 2005 #4
    In essence all you have to do is ensure that your Jacobian is injective (ie: one-to-one) and that its determinant is non-zero.
  6. Dec 16, 2005 #5
    Ok, but sometimes it doesn't seem so easy to decide whether or not the transformation is one to one. For example in this question I used the substitution u = sinhxcosy.

    To decide whether the transformation is one to one, is it like the on variable integral where you have something like u = x + 1, 0<=x<=2 and you determine whether that substitution (transformation) is one to one in terms of a graph. Or, do you need to take a matrix and use it's Kernel or something like that to determine whether the transformation is one to one?
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