How do I calculate a double integral like $\iint_\mathbf{D}e^{\frac{x-y}{x+y}}dx\,dy$? The Problem is to integrate the following double integral $\displaystyle\iint_\mathbf{D}\exp\left(\frac{x-y}{x+y}\right) dx\, dy$ using the technique of transformation of variables ($u$ and $v$).
There is a given $D$ with $D:= \{(x,y):\ x\geq 0,\ y \geq 0,\ x+y \leq 1\}$
My Approach: wasn't quite successful at all. I am quite familar with calculus, but i am stuck at multiple integrals. I thought about the following transformation:  $u = x-y$ and $v = x+y$ 
But i don't know the next step.
Another suggestion: i thought about the integration of the e-Function. I have in mind that $\int e^x dx = e^x$. So maybe: $$\iint_\mathbf{D} \exp\left(\frac{x-y}{x+y}\right) dx\ dy = \exp\left(\frac{x-y}{x+y}\right) ?$$ 
 A: Consider the transformation:
$u=x+y$ and $v=\dfrac{x}{x+y}$
$\implies x=uv$ and $y=u(1-v)$.
Here our domain is $D= \{(x,y):\ x\geq 0,\ y \geq 0,\ x+y \leq 1\}$
As $ x\geq 0,\ y \geq 0,\ x+y \leq 1\\\implies  uv\geq 0,\ u(1-v) \geq 0,\ u \leq 1\\\implies  0\leq u\leq 1,0\leq v \leq 1$
The transformed domain is $E=\{(u,v):0\leq u\leq 1,0\leq v \leq 1\}$
Again the jacobian id $J=-u$
Hence $\int_\mathbf{D}\int e^{\dfrac{x-y}{x+y}}\, dx\, dy\\=\int_0^1\int_0^1e^{\dfrac{uv-u(1-v)}{u}}|-u|\, du\, dv\\=(\int_0^1u\,du)( \int_0^1e^{2v-1}\,dv) $
A: With your substitution, $x=\frac{1}{2}(u+v)$ and $y=\frac{1}{2}(v-u)$.  Hence $D$ becomes $$\left\{(u,v):\frac{1}{2}(u+v)\ge 0, \frac{1}{2}(v-u)\ge 0, v\le 1\right\}$$
To switch from $dxdy$ to $dudv$ you pay the price of the determinant of the Jacobian, which is $$\left|\begin{smallmatrix} 0.5& 0.5\\-0.5&0.5\end{smallmatrix}\right|=0.5$$
A: It's a bit underhanded, admittedly, but you can also do this problem in polar coordinates. You will get
$$\int_0^{\pi/2}\int_0^{1/(\cos\theta+\sin\theta)} e^{\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}} r\,dr\,d\theta\,,$$
which looks totally horrifying. However, the $r$ integral provides you a factor of $\dfrac1{(\cos\theta+\sin\theta)^2}$, which just perfectly makes the $\theta$ integral manageable.
