# Calculate integral through a change of variables

1. Jul 12, 2013

### tsuwal

1. The problem statement, all variables and given/known data

Let
$D=\{ (x,y)\in\mathbb{R}^2:x+y< 1;0< y< x\}$
calculate $\int_{D} e^{-(x+y)^4}(x^2-y^2)$through an appropriate change of variables

2. Relevant equations

$\int_{D} f *dxdy=\int_{D} f*Jacobian*dudv$

3. The attempt at a solution

I've tried x+y=u and x-y=v which is a linear transformation so the jacobian is constant. However the integral becomes pretty ugly meanwhile and I can't solve it. What do i do?

2. Jul 12, 2013

### lurflurf

Try something more like

u=x+y
v=x

3. Jul 12, 2013

### HallsofIvy

Staff Emeritus
On the contrary, with that substitution, the integral becomes very nice! With u= x+ y. $e^{-(x+y)^4}= e^{u^4}$ and with, also, v= x- y, $x^2- y^2= (x- y)(x+ y)= uv$ so the integrand becomes $e^{u^4}uv$. Of course, the boundaries, x+y= 1, y= 0, and y= x, in the u, v system, become u= 1, v= 0, and u= v. The integral is
$$2\int_0^1\int_0^u e^{-u^4}uv dvdu$$

Now, do the first, very easy, integral with respect to v and then slap your forehead and cry "Of course"!

4. Jul 12, 2013

### tsuwal

yeah right, did a mistake on the limits of integration of the v variable. I calculated the limiting points of the triangle in the uv plane and did a mistake in one point. Limits of v became v=u and v=1. This way was impossible.

Thanks but I think you did a mistake on the Jacobian. It is 1/2 not 2.