Considering an appropriate lineat mapping $(x,y)=T(u,v)$, I want to compute by changing the variable the integral $\iint_D e^{\frac{y-x}{y+x}}dxdy$, where $D$ os the triangle with edges the point $(0,0)$, $(1,0)$ and $(0,1)$.
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I have done the following:
We define the new variables $u=x+ y$, $v=x-y$.
When we solve for $x,y$ as a function of $u,v$ we get $$x=\frac{u}{2}+\frac{v}{2}, \ y=\frac{u}{2}-\frac{v}{2}$$
So we get the following linear mapping $$\binom{x}{y}=T\left (\binom{u}{v}\right )=\begin{pmatrix}\frac{u}{2}+\frac{v}{2} \\ \frac{u}{2}-\frac{v}{2}\end{pmatrix}=\begin{pmatrix}\frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & -\frac{1}{2}\end{pmatrix}\binom{u}{v}$$
At which intervals do the new variables lie?
To calculate the bounds for $u$, $v$ you should map the area $D$ to the new coordinate system $uv$. Since the mapping is linear, the lines map to lines, so it is enough to map just the vertices of the triangle and to get the new triangle as $(0,0)$, $(1,1)$ and $(1,-1)$.