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Clarification of change of variables for multiple integration

  1. Jul 14, 2009 #1
    Clarification of "change of variables" for multiple integration

    This isn't really a question about a specific math problem, but rather for the change of variables of multiple integration as a whole. When you change variables you have to multiply the new expression by the jacobian of the new functions you chose. So, if the determinant is a positive constant, you can just bring it outside of the integral signs. However, in my book, whenever the determinant is a negative constant, they bring out its positive reciprocal instead (For example, if the jacobian determinant is -1/3, they bring out a 3 instead)
    Is this the proper way to do it?
    For example, there is q eustion where you use the substituion u = x+y, v = x-y
    the determinant of this is -2 ( ad - bc = -1 - 1 = 2), so in the solutions in the back of the book they've turned they've multiplied the new expression by 1/2 instead of by -2

    Is this the rule? If determinant is a positive constant, you use that constant, and if it's a negative one, you use its positive reciprocal?
     
  2. jcsd
  3. Jul 14, 2009 #2
    Re: Clarification of "change of variables" for multiple integration

    I think I can see where your confusion is on this one. It is admirable of you to look for such new rules/hidden patterns, but it is much simpler than this. There are two things that might make you think that there is some strange "negative reciprocal" rule when, in fact, there isn't:

    First of all, when transforming from one coordinate system to another, you always use the absolute value of the Jacobian. This accounts for why it looked like a negative was turning into a positive.

    Secondly, you have the wrong version of your transformation. I'm guessing that your original problem has you integrating over the xy-plane and then telling you to switch to the uv-plane using the change of variables:
    u = x + y
    v = x-y
    right? If this is the case, then you need to invert those equations before you compute the Jacobian. Remember, the Jacobian matrix is formed by differentiating the old variables with respect to the new ones. Your equations give you the exact opposite: the new variables with respect to the old ones. If you invert them (a fairly easy exercise in this case) by solving for x and y you get:
    x = 1/2(u+v)
    y = 1/2(u-v)
    Compute the Jacobian of this and you will now get -1/2 instead of -2.

    You can see that this reciprocal relation for the Jacobian will hold for at least all linear transformations since:
    [tex]AA^{-1}=a[/tex]
    and, taking the determinant of both sides we find
    [tex]\det{AA^{-1}}=\det{A}\det{A^{-1}}=1[/tex]
    so that
    [tex]\det{A^{-1}}=\frac{1}{\det{A}}[/tex]
    And as long as the transformation of variables is linear, the Jacobian will always be constant, so it can be pulled outside of the integral. I'm not sure about the nonlinear case, but I doubt the reciprocal nature still holds.

    Hopefully that helps!
     
  4. Jul 14, 2009 #3
    Re: Clarification of "change of variables" for multiple integration

    Damn, I just wrote my test and I differentiated all the new variables with respect to the old ones instead of the other way around :( damnit!

    But I now understand why this work :) And this definitely solves my problem because I redid a few exercises and got the right answers now. Thank you very much!
     
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