Evaluate $\iint_{D}e^{(x+y)^2}dxdy$, where $D$ is the interior of the triangle with vertices $(0,0), (3,1)$ and $(2,2)$ I need to solve this problem for my vector calculus class:

Evaluate $\iint_{D}e^{(x+y)^2}\,dxdy$, where $D$ is the interior of the triangle with vertices $(0,0), (3,1)$ and $(2,2)$.

I first tried using polar coordinates but I ended up with an unsolvable integral so I concluded this wasn't the right way to do it.
I then tried to find a map $T(u,v)$ but I couldn't find one that could help me solve this integral. I tried $x=u+v$ and $y=u-v$ but $\frac{\partial (x,y)}{\partial(u,v)}=2$ and I get $\iint_{D}2e^{4u^2}dudv$ which is not solvable using classic functions.
Any ideas on what change of variable should I use?
Thank you in advance for your answers
 A: You could try the substitution $x+y = \sqrt s$, $x-y = t$.  The region in question has $x+y \ge 0$ so this is well-defined and transformation from $(s,t)$ space to $(x,y)$ space given by
\begin{align*} x&= \frac 12(\sqrt s + t) \\
y&= \frac 12(\sqrt s - t)\end{align*} is one-to-one.
The equations of the sides of the triangle are $x=y$, $x+y = 4$, and $x=3y$.  These correspond to the equations $t = 0$, $s = 16$, and $s = 4t^2$ in $(s,t)$ space.  Thus your triangle $T$ is obtained by transforming the region $D$ in the first quadrant of $(s,t)$ space bounded below by $t = 0$, bounded on the right by $s = 16$, and bounded above by $t = \frac {\sqrt{s}}{2}$.
The Jacobian of the transformation is easily computed to be
$$ \frac{\partial(x,y)}{\partial (s,t)} = - \frac{1}{4 \sqrt s}$$ and so the change of variable formula tells you
$$\int_T e^{(x+y)^2}\,dydx = \int_D e^s \left| \frac{\partial(x,y)}{\partial (s,t)} \right| \, dtds = \int_0^{16} \int_0^{\sqrt s/2} e^s \frac{1}{4 \sqrt s} \, dtds = \frac 1{8} \int_0^{16} e^s \, ds$$ which evaluates to $$ \frac 1{8} (e^{16} - 1).$$
A: Once you make the substitution
$$
\left\{ \begin{array}{l}
 x = u + v \\ 
 y = u - v \\ 
 \end{array} \right.\quad  \Leftrightarrow \quad \left\{ \begin{array}{l}
 u = \frac{{x + y}}{2} \\ 
 v = \frac{{x - y}}{2} \\ 
 \end{array} \right.
$$
then the triangle becomes
$$
T_{x,y}  = \left\{ {\left( {0,0} \right),\left( {3,1} \right),\left( {2,2} \right)} \right\}\quad  \Leftrightarrow
 \quad T_{x,y}  = \left\{ {\left( {0,0} \right),\left( {2,1} \right),\left( {2,0} \right)} \right\}
$$
and since the integrand depends only on $u$ it is easy to separate it, integrate first on $v$ and ...
A: We can switch to polar coordinates as the following picture shows:
(I didn't know how to make picture smaller :)















First, observe that $$x+y = 4 \implies r = \frac{4}{\cos\theta + \sin\theta}$$
From the picture, we have $$r \in \left[0, \frac{4}{\cos\theta + \sin\theta}\right]$$ and $$\theta \in \left[\arctan\frac{1}{3}, \frac{\pi}{4}\right]$$ Now, the integral should be easier.
