How should I solve this integral with changing parameters?

Question

I can't solve this. How should I proceed?

$$\iint_De^{\large\frac{y-x}{y+x}}\mathrm dx\mathrm dy$$

$D$ is the triangle with these coordinates $(0,0), (0,2), (2,0)$ and I've changed the parameters this way $u=y-x$ and $v= y+x$ and the Jacobian is $-\frac{1}{2}$ but I have problem finding the range of $u$ and $v$ to calculate the integral.

Answer 1

Start by making a drawing of your domain. You can see that the $v$ is along the diagonal in the first quadrant, and $u$ is along the diagonal in the second quadrant. You can also see that the line between $(0,2)$ and $(2,0)$ is parallel to $u$, and intersects $v$ axis at $v=1$. So $v$ varies between $0$ and $1$ and $u$ varies between $-v$ and $v$.

Answer 2

Computer algebra gives (for the general case):

$$e \left(x^2 \text{Ei}\left(-\frac{2 x}{x+y}\right)-\frac{y^2 \text{Ei}\left(\frac{2 y}{x+y}\right)}{e^2}\right)+\frac{1}{2} e^{1-\frac{2 x}{x+y}} (x+y)^2$$

over your specified region:

$$e-\frac{1}{e}$$

Answer 3

Your domain is $$D=\{(x,y):0<x<2\,,\,0<y<2\,,\,x+y<2\}$$

Using change of variables $(x,y)\to(u,v)$ with $$u=\frac{x-y}{x+y}\,,\,v=x+y$$

The region is now $$R=\{(u,v):-1<u<1\,,0<v<2\}$$

Therefore,

\begin{align} \iint_D \exp\left({-\frac{x-y}{x+y}}\right)\,\mathrm{d}x\,\mathrm{d}y&=\frac{1}{2}\iint_R ve^{-u}\,\mathrm{d}u\,\mathrm{d}v \\\\&=\frac{1}{2}\int_0^2v\,\mathrm{d}v\int_{-1}^1 e^{-u}\,\mathrm{d}u \end{align}