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Math Help - Integration of e with Jacobian

  1. #1
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    Integration of e with Jacobian

    Hi, could someone please explain how I integrate this?

    \int \int e^(-g(3x-4y)^4)dxdy

    where g is a positive constant

    I have a Jacobian of -3, from
    x = 2u + v; y =u - v

    and the region is described by x is less than or equal to 0 and y is greater than or equal to 0
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  2. #2
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    Quote Originally Posted by Penguin91 View Post
    Hi, could someone please explain how I integrate this?

    \int \int e^(-g(3x-4y)^4)dxdy

    where g is a positive constant

    I have a Jacobian of -3, from
    x = 2u + v; y =u - v

    and the region is described by x is less than or equal to 0 and y is greater than or equal to 0
    If x = 2u + v and y = u - v

    Then J = \left[\begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{matrix}\right]

     = \left[\begin{matrix}2 & 1 \\ 1 & -1\end{matrix}\right]


    |J| = 2(-1) - 1(1) = -2 - 1 = -3.


    So the change of variable is dx\,dy = -3\,du\,dv.


    So e^{-g(3x - 4y)^4}\,dx\,dy = e^{-g[3(2u + v) - 4(u - v)]^4}\,-3\,du\,dv

     = -3e^{-g(2u + 7v)^4}\,du\,dv.


    Now just change your boundaries and you will be ready to go.
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  3. #3
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    Thanks for your reply, I'm a bit new to this topic so I'm still slightly confused, when I try to change the limits, I get -v/2 <= u <= -v/2 for x
    and for y u <= v <= u, how does this work? :S
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