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Double Integration: Can I finish?

  1. Feb 22, 2012 #1

    Char. Limit

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    1. The problem statement, all variables and given/known data
    Evaluate the integral by making an appropriate change of variable: [itex]\int \int_R (x + y) e^{x^2 - y^2} dA[/itex], where R is the rectangle enclosed by the lines x-y=0, x-y=2, x+y=0, and x+y=3

    2. Relevant equations
    If x=x(u,v) and y=y(u,v), then [itex]\int \int f(x,y) dx dy = \int \int g(u,v) |J(u,v)| du dv[/itex], where J(u,v) is the Jacobian determinant of x and y wrt u and v.

    3. The attempt at a solution

    So, I begin by selecting my u and v as u=x+y and v=x-y, which gives my bounds as a nice rectangle: 0<v<2 and 0<u<3. With that, I get my new function as:

    [tex]\int_0^2 \int_0^3 u e^{u v} |J(u,v)| du dv[/tex]

    Finding the Jacobian, I get it to be -1/2, and thus |J(u,v)| = 1/2. Okay, so far so good. I take my first integration:

    [tex]\frac{1}{2} \int_0^2 \left(\frac{u}{v} e^{u v} - \frac{1}{v^2} e^{u v}\right)|_0^3 dv = \frac{1}{2} \int_0^2 \frac{3 e^{3v}}{v} - \frac{e^{3v}}{v^2} + \frac{1}{v^2} dv[/tex]

    Now, at this point I realized I should have changed the order of integration. However, I'm wondering if I can still find this integral. The first two terms look suspiciously like the derivative of e^(3v)/v, after all... so my question is, is it still easy to continue from here, or should I just start all over and change the order of integration?
  2. jcsd
  3. Feb 23, 2012 #2
    Yes, the first two terms are the derivative of [itex]\displaystyle\frac{e^{3v}}{v}[/itex]. Rewrite them as such and simply use the Fundamental Theorems of Calculus. It is find the way it is.
  4. Feb 23, 2012 #3

    Char. Limit

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    Excellent! I finished the second half of the problem and got (e^6 - 7)/4 as my answer. Danke!
  5. Feb 23, 2012 #4


    Staff: Mentor

    From your sig:
    Yeah, right...:smile:
  6. Feb 23, 2012 #5

    Char. Limit

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    But it's true! Albeit indirectly and not from PF.
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