Jacobian Integration Problem

**llamagoogle** · February 23rd 2010, 10:41 AM

$\int \int e^(-r(x-2y)^2) dxdy$

x =2u+v
y=u-v

The Jacobian is -3
I've substituted for x and y in the integral to get $\int \int -3e^r(3v)^2 dudv$

where r is a positive constant

But my problem lies with the limits, I have y is greater than or equal to 0 and x is less than or equal to 0, How do I change these for the integral and proceed as a result?

Thanks

**tonio** · February 23rd 2010, 11:00 AM

Originally Posted by llamagoogle

$\int \int e^(-r(x-2y)^2) dxdy$

x =2u+v
y=u-v

The Jacobian is -3
I've substituted for x and y in the integral to get $\int \int -3e^r(3v)^2 dudv$

where r is a positive constant

But my problem lies with the limits, I have y is greater than or equal to 0 and x is less than or equal to 0, How do I change these for the integral and proceed as a result?

Thanks

What you wrote in the 1st integral doesn't show clearly. Anyway:

$\left\{\begin{array}{c}x=2u+v\\y=\;\;u-v\end{array}\right.$ $\Longrightarrow x+y=3u\Longrightarrow u=\frac{x+y}{3}\Longrightarrow v=\frac{x-2y}{3}$ , so $-\infty < u<\infty\,,\,\,-\infty<v\leq 0$

Tonio

**Moo** · February 23rd 2010, 11:02 AM

Hello,

Hmmm the Jacobian has to be the absolute value of the determinant, so it would be 3.

now for the boundaries :

we have $x=2u+v\leq 0$ and $y=u-v\geq 0$

from this, it follows that $\boxed{v\leq u\leq -\tfrac v2}$

and for the boundaries of v : $v\leq -\tfrac v2 \Rightarrow \boxed{v\leq 0}$

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