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Math Help - Double integral change of variables

  1. #1
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    Double integral change of variables

    I have to integrate x+y with 0<=x<=1 and 0<=y<=x by making the change of variables x=u+v and y=u-v.

    I got the Jacobian to be 2, and I solved for u and v and got u=(x+y)/2 and v=(x-y)/2, but I can't seem to find the right limits of integration. Any hints?
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  2. #2
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    Re: Double integral change of variables

    Your region of integration is bound by the triangle whose vertices are at (0,0), (1,1), and (1,0).

    If x=0 & y=0,the (u, v) = (0, 0). If x=1 & y=1, then (u, v) = (1, 0). If x=1 & y=0, then u = 1/2 & v = 1/2 .

    Integrate 0 ≤ u ≤ 1, 0 ≤ v ≤ (1/2) - |x - (1/2)|
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    Re: Double integral change of variables

    Thanks! I'm also having trouble with this one.

    Let R denote the region inside x^2 + y^2 = 1 but outside x^2 + y^2 = 2y with x=>0 and y=>0. Let u=x^2 + y^2 and v=x^2 + y^2 -2y. Compute the integral of x*e^y over the region D in the uv-plane which corresponds to R under the specified change of coordinates.

    My first attempt at figuring out the new limits of integration yielded 0<=u<=1 and 0<=v<=u, which seems wrong to me. I'm also not sure how to change the integrand to make it a function of u and v.
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  4. #4
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    Re: Double integral change of variables

    Quote Originally Posted by SammyS View Post
    Your region of integration is bound by the triangle whose vertices are at (0,0), (1,1), and (1,0).

    If x=0 & y=0,the (u, v) = (0, 0). If x=1 & y=1, then (u, v) = (1, 0). If x=1 & y=0, then u = 1/2 & v = 1/2 .

    Integrate 0 ≤ u ≤ 1, 0 ≤ v ≤ (1/2) - |u - (1/2)|
    Added in Edit:

    Switch order of integration. It makes more sense.

    Integrate: 0 ≤ v ≤ 1/2 , v ≤ u ≤ 1-v
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  5. #5
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    Re: Double integral change of variables

    What about the second problem in post #3?
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  6. #6
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    Re: Double integral change of variables

    Yup. That's a lot tougher.

    Have you posted that elsewhere?
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  7. #7
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    Re: Double integral change of variables

    yes, but still no solution.
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