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Math Help - change of var, jacobian

  1. #1
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    change of var, jacobian

    Hi Guys,
    I'm having troubles starting this problem. In the previous problems we have had, our transformation has been given to us (i.e something like x=u+v,y=u-v etc...). Since one is not given here do I just use x as u and y as v? Any help with this nasty would be greatly appreciated.


    Evaluate the integral where R is the rectangle enclosed by the lines
    x - y = 0,
    x - y = 2,
    x + y = 0,
    x + y = 3.
    Use a change of variable and a Jacobian.
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  2. #2
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    Note that all your limits involve x+y or x-y, and that part of your integrand is (x+y). Does this suggest a choice for u and v?

    --Kevin C.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by boousaf View Post
    Hi Guys,
    I'm having troubles starting this problem. In the previous problems we have had, our transformation has been given to us (i.e something like x=u+v,y=u-v etc...). Since one is not given here do I just use x as u and y as v? Any help with this nasty would be greatly appreciated.


    Evaluate the integral where R is the rectangle enclosed by the lines
    x - y = 0,
    x - y = 2,
    x + y = 0,
    x + y = 3.
    Use a change of variable and a Jacobian.
    \iint\limits_R \left(x+y\right)e^{x^2+y^2}\,dA

    Where

    x - y = 0,    \ x - y = 2,   \ x + y = 0, \text{ and } x + y = 3

    So letting u=x-y and v=x+y, we see that we have the limits 0\leq u\leq 2 and 0\leq v\leq 3

    Now we need to find equations for x and y in terms of u and v.

    since u=x-y \text{ and } v=x+y, we see that u+y=x \text{ and }v-y=x

    Therefore, u-y=v+y\implies \boxed{y=\tfrac{1}{2}(u-v)}

    Now let's find an equation for x.

    since u=x-y \text{ and } v=x+y, we see that x-u=y \text{ and }y=v-x

    Therefore, x-u=v-x\implies \boxed{x=\tfrac{1}{2}(u+v)}

    Now, let's find the Jacobian J(u,v).

    J(u,v)=\frac{\partial(x,y)}{\partial(u,v)}=\bigg|\  begin{array}{cc}\tfrac{\partial x}{\partial u} & \tfrac{\partial x}{\partial v} \\ \tfrac{\partial y}{\partial u} & \tfrac{\partial y}{\partial v}\end{array}\bigg|=\bigg|\begin{array}{cc}\tfrac{  1}{2} & \tfrac{1}{2} \\ \tfrac{1}{2} & -\tfrac{1}{2}\end{array}\bigg|=-\tfrac{1}{2}

    Now change the variables of the integral:

    \iint\limits_R \left(x+y\right)e^{x^2+y^2}\,dA\implies \tfrac{1}{2}\iint\limits_{R_{uv}} ve^{\tfrac{1}{2}(u^2+v^2)}\,dA_{uv}\implies \tfrac{1}{2}\int_0^2\int_0^3 ve^{\tfrac{1}{2}(u^2+v^2)}\,dv\,du

    Can you try to integrate this? I think you can take it from here.

    --Chris
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  4. #4
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    thanks chris! a wonderful explanation
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  5. #5
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    The Jacobian is positive, it's actually y=\frac{v-u}2.
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  6. #6
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Krizalid View Post
    The Jacobian is positive, it's actually y=\frac{v-u}2.
    I figured that I would miss a sign somewhere...

    --Chris
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