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Change of Variables of f(x+y) (In Multiple Integrals)

  1. Nov 29, 2009 #1
    1. The problem statement, all variables and given/known data
    The problem is as follows: Let T be the triangle with vertices (0,1), (1,0), (0,0). Compute the integral [tex]\int\int[/tex][tex]\frac{sin^{2}(x+y)}{(x+y)} dxdy[/tex] by making an appropriate change of variables. (Hint: check #24 Section 15.9)


    2. Relevant equations
    Problem 24 in 15.9 of Stewart Calculus Early Transcendentals: Let f be continuous on [0,1] and let R be the triangular region with vertices (0,1), (1,0) and (0,0). Show that
    [tex]\int\intf(x+y)dA[/tex] [tex]=\int_{0}^{1}uf(u)du[/tex]



    3. The attempt at a solution
    I am confused at what values I should assign u and v in order to change the variables appropriately. Assuming the answer to #24, I obtained the solution to be (1/2)-(1/4)sin(2)
    Thanks for any help or tips that point me in the right direction,
    Jim
     
  2. jcsd
  3. Nov 29, 2009 #2

    LCKurtz

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    The key to #24 is the fact that the side of the triangle not on one of the axes has the equation x + y = 1. That suggests letting one of the variables by x + y. And there is no reason to change the other variable, so try:

    u = x + y
    v = x

    The corners of the xy triangle go to (u,v) like this:

    (0,0 ) --> (0,0)
    (1,0) --> (1,1)
    (0,1) --> (1,0)

    Your exercise is to show that

    [tex]\int\int_A f(x+y)\,dA = \int_0^1uf(u)\,du[/tex]

    Notice on the right side, what would normally be a double integral has been integrated in the v variable (the one that isn't in the f()) already, so set up the change of variables that way. The way we set it up, the non-x+y variable is v also. Since |J| = 1 I will leave the Jacobian out. Our vu limits become:

    [tex]\int_0^1\int_0^u f(u)\,dvdu[/tex]

    Now you can do the inner integral without knowing f.
     
  4. Nov 29, 2009 #3
    Got it, thanks!
     
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