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Calculate integral through a change of variables

  1. Jul 12, 2013 #1
    1. The problem statement, all variables and given/known data

    Let
    [itex]
    D=\{ (x,y)\in\mathbb{R}^2:x+y< 1;0< y< x\}
    [/itex]
    calculate [itex] \int_{D} e^{-(x+y)^4}(x^2-y^2) [/itex]through an appropriate change of variables


    2. Relevant equations

    [itex] \int_{D} f *dxdy=\int_{D} f*Jacobian*dudv [/itex]

    3. The attempt at a solution

    I've tried x+y=u and x-y=v which is a linear transformation so the jacobian is constant. However the integral becomes pretty ugly meanwhile and I can't solve it. What do i do?
     
  2. jcsd
  3. Jul 12, 2013 #2

    lurflurf

    User Avatar
    Homework Helper

    Try something more like

    u=x+y
    v=x
     
  4. Jul 12, 2013 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    On the contrary, with that substitution, the integral becomes very nice! With u= x+ y. [itex]e^{-(x+y)^4}= e^{u^4}[/itex] and with, also, v= x- y, [itex]x^2- y^2= (x- y)(x+ y)= uv[/itex] so the integrand becomes [itex]e^{u^4}uv[/itex]. Of course, the boundaries, x+y= 1, y= 0, and y= x, in the u, v system, become u= 1, v= 0, and u= v. The integral is
    [tex]2\int_0^1\int_0^u e^{-u^4}uv dvdu[/tex]

    Now, do the first, very easy, integral with respect to v and then slap your forehead and cry "Of course"!
     
  5. Jul 12, 2013 #4
    yeah right, did a mistake on the limits of integration of the v variable. I calculated the limiting points of the triangle in the uv plane and did a mistake in one point. Limits of v became v=u and v=1. This way was impossible.

    Thanks but I think you did a mistake on the Jacobian. It is 1/2 not 2.
     
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