# Math Help - Integration of e with Jacobian

1. ## 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

2. Originally Posted by Penguin91
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.

3. 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