<s>[INST] Creating my own change of variables to evaluate an integral The question asks me to evaluate the integral $$\iint_{R} e^{\frac{x+y}{x-y}} dA$$ where $R$ is trapezoid region with the vertices $(1,0), (2,0), (0,-2), (0,-1)$. I'm supposed to suggest a possible transformation and integrate and sketch the two regions.
My work :
Let the transformation be
$u=x-y$, $v=x+y$
Then with some algebra, I get $x=\frac{u+v}{2}$, and $y=-\frac{1}{2} (u-v)$
$J(u,v)=\begin{vmatrix}
\frac{1}{2} & \frac{1}{2}\\ 
-\frac{1}{2} & \frac{1}{2}
\end{vmatrix}=\frac{1}{2}$
When I sketch the region I have something like this on the xy plane

On the uv plane the transformation looks like:

So the integral becomes
$$\int_{1}^{2}\int_{-u}^{u} e^{\frac{v}{u}}*\frac{1}{2} dv du$$
$$\frac{1}{2}\int_{1}^{2}u\Big(e-\frac{1}{e}\Big)du$$
$$=\frac{1}{2}\Big(e-\frac{1}{e}\Big)*\frac{3}{2}=\frac{3}{4}\Big(e-\frac{1}{e}\Big)$$
Does this look correct?
 [/INST] Your approach is nice and the calculation is well done. We can check it, by calculating it slightly different and look if the results coincide. We apply the identity
\begin{align*}
\frac{x+y}{x-y}=1+\frac{2y}{x-y}
\end{align*}
and consider
\begin{align*}
\iint_{R} e^{\frac{x+y}{x-y}} dA=e\iint_{R} e^{\frac{2y}{x-y}} dA\tag{1}
\end{align*}
We use the variable transformation
\begin{align*}
u&=y\qquad\qquad\qquad x=u+v\\
v&=x-y\qquad\qquad\ y=u
\end{align*}
The trapezoid regions have vertices
\begin{align*}
\mathrm{Tr}_{(x,y)}&=\{(0,-2),(0,-1),(1,0),(2,0)\}\\
\mathrm{Tr}_{(u,v)}&=\{(-2,2),(-1,1),(0,1),(0,2)\}\\
\end{align*}
and the transformed region is given by the graphic below.

The Jacobian determinant is
\begin{align*}
J(u,v)=
\begin{vmatrix}
x_u&x_v\\
y_y&y_v\\
\end{vmatrix}
=
\begin{vmatrix}
1&1\\
1&0
\end{vmatrix}=-1
\end{align*}

We obtain with the right-hand side of (1)
\begin{align*}
\color{blue}{e\iint_{R} e^{\frac{2y}{x-y}} dA}
&=e\int_1^2\int_{-v}^0e^{\frac{2u}{v}}|-1|\,du\,dv\\
&=e\int_{1}^2\left.\frac{v}{2}e^{\frac{2u}{v}}\right|_{-v}^0\,dv\\
&=\frac{e}{2}\int_{1}^2v\left(1-e^{-2}\right)\,dv\\
&=\frac{1}{2}\left(e-\frac{1}{e}\right)\left.\frac{1}{2}v^2\right|_{1}^2\\
&\,\,\color{blue}{=\frac{3}{4}\left(e-\frac{1}{e}\right)}
\end{align*}
in accordance with OPs result.

</s>