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Math Help - Using appropriate transformation for integration..

  1. #1
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    Using appropriate transformation for integration..

    Can anyone help with how to attack this question? thanks!
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by calciihelp View Post
    Can anyone help with how to attack this question? thanks!
    Make the substitutions u=x^2+y^2 and v=x^2-y^2. (Also keep in mind that with these substitutions, 4\leq u\leq9 and 1\leq v\leq2)

    It follows then that u-x^2=y^2 and x^2-v=y^2

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

    Similarly, u-y^2=x^2 and v+y^2=x^2

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

    Now, we need to find the Jacobian:

    J(u,v)=\begin{vmatrix}\displaystyle\frac{\partial x}{\partial u} &\displaystyle\frac{\partial x}{\partial v}\\\displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v}\end{vmatrix}=\begin{vmatrix}\displaystyle\frac{  1}{2\sqrt{2(u+v)}} & \displaystyle\frac{1}{2\sqrt{2(u+v)}}\\\displaysty  le\frac{1}{2\sqrt{2(u-v)}}&\displaystyle-\frac{1}{2\sqrt{2(u-v)}}\end{vmatrix} =-\frac{1}{8\sqrt{u^2-v^2}}-\frac{1}{8\sqrt{u^2-v^2}}=-\frac{1}{4\sqrt{u^2-v^2}}

    Therefore, \iint\limits_{S}xy\,dA\implies\int_1^2\int_4^9\sqr  t{\tfrac{1}{2}(u+v)}\sqrt{\tfrac{1}{2}(u-v)}\cdot\left|-\frac{1}{4\sqrt{u^2-v^2}}\right|\,du\,dv=\tfrac{1}{8}\int_1^2\int_4^9\  ,du\,dv

    (Since we're looking for the area in the first quadrant, we might have to divide our answer by 4).

    I'm sure you can take it from here...
    Last edited by Chris L T521; August 3rd 2009 at 12:30 PM. Reason: fixed mistake
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  3. #3
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    Those circles are represented by 4\le x^2+y^2\le9 so 4\le u\le9.

    Besides, use

    \frac{\partial (x,y)}{\partial (u,v)}=\frac{1}{\dfrac{\partial (u,v)}{\partial (x,y)}},

    so there's no need to isolate variables.
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