Jacobian: int[0 to 2/3] int[y to 2-2y] (x+2y)(e)^(y-x) dxdy

mammothrob

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Nov 12, 2005
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I have this jacobian problem. It's a double integral and the integrand is

[0 to 2/3] [y to 2-2y] (x+2y)(e)^(y-x) dxdy

My problem says to use u=x+2y and v=x-y

Notice v doesnt match into the integral right
is there a typo in the problem because should'nt my v=y-x instead so it will fit into the integral nicely?

If anyone sees how to do this the way it is could, can you show me how to get the new region with the transformations.

I know that the orginal is a triangle, but that trans forms that I use don't make a tiangle? no solid is created
 
\(\displaystyle \L\\\int\int(x+2y)(e^{y-x})dxdy\)

\(\displaystyle x=y; \;\ x=2-2y; \;\ y=0; \;\ y=\frac{2}{3}\)

Using what they suggest, I end up getting:

\(\displaystyle \L\\y=\frac{u-v}{3}\)

\(\displaystyle \L\\x=\frac{u+2v}{3}\)

Finding the partial derivative and the determinant,

This transforms the integral into:

\(\displaystyle \L\\\frac{1}{3}\int\int{ue^{-v}}dudv\)

Check it out. Thi seems on the right track.

Can you find the limits of integration for the transform now?.
 
Hey thanks. I was doing something really weird. I got the new region also and limits. It worked out really well. Thanks again.
Rob
 
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