The easiest way to evaluate $I=\iint_D e^{-(x^2+y^2)}\,dx\,dy,\ \ \ D=\left\{(x,y)\Bigm|2\leqslant |x|+|y|\leqslant 3\right\}$ 
Evaluate the following integral:
  $$
I=\iint_D e^{-(x^2+y^2)}\,dx\,dy,\ \ \ D=\left\{(x,y)\Bigm|2\leqslant |x|+|y|\leqslant 3\right\}
$$

Well, to make things a bit easier, I can find an integral $J$ first:
$$
J=\int\limits_0^3dx\int\limits_0^{-x+3}e^{-x^2-y^2}dy
$$
And then
$
I=4J
$.
Afterwards, if I try to make a polar coordinate substitution, I get some problems:
$$
J=\int\limits_0^{\pi/2}d\phi\int\limits_0^{\frac{3}{\sin\phi+\cos\phi}}e^{-r^2}r\,dr=\frac{1}{2}\int\limits_0^{\pi/2}\left(e^{-\left(\frac{3}{\sin\phi+\cos\phi}\right)^2}-1\right)d\phi
$$
And it is rather difficult to evaluate the final integral. So, perhaps there is a better approach to this problem?
 A: Let $u=\frac{x+y}{\sqrt2}$ and $v=\frac{y-x}{\sqrt2}$.  Then the $(u,v)$ coordinates $(u,v)$ are obtained from $(x,y)$ by a $45^\circ$-rotation in the clockwise direction. (In my comment above, I mistakenly defined $v=\frac{x-y}{\sqrt2}$.  Call this $v'$ instead to avoid confusion.  While most of the idea remains unchanged, it should be noted that in the comment, the $(u,v')$ coordinates is obtained by a rotation and a reflection of the $(x,y)$ coordinates.)
Then in the new coordinates, the image $D'$ of $D$ looks like
$$D'=\left\{(u,v)\in\Bbb{R}^2:\frac{2}{\sqrt2}\le |u|,|v|\le \frac{3}{\sqrt2}\right\}=A\setminus B,$$
where $A$ and $B$ are the squares
$$A=\left\{(u,v)\in\Bbb{R}^2:|u|,|v|\le \frac{3}{\sqrt2}\right\}$$
and $$B=\left\{(u,v)\in\Bbb{R}^2:|u|,|v|< \frac{2}{\sqrt2}\right\}.$$
Since $u^2+v^2=x^2+y^2$, we get
$$\iint_D e^{-x^2-y^2} dx\ dy=\iint_{D'}e^{-u^2-v^2}du\ dv =\iint_A e^{-u^2-v^2}du\ dv-\iint_B e^{-u^2-v^2}du\ dv.$$
Clearly
\begin{align}\iint_A e^{-u^2-v^2}du\ dv&=\int_{-3/\sqrt2}^{3/\sqrt2} \int_{-3/\sqrt2}^{3/\sqrt2}  e^{-u^2-v^2}du\ dv\\&=\int_{-3/\sqrt2}^{3/\sqrt2}e^{-u^2}du \int_{-3/\sqrt2}^{3/\sqrt2}  e^{-v^2}dv\\&=\left(\int_{-3/\sqrt2}^{3/\sqrt2}e^{-t^2}dt\right)^2=\Biggl(\sqrt{\pi}\operatorname{erf}\left(\frac{3}{\sqrt2}\right)\Biggr)^2.\end{align}
Likewise
\begin{align}\iint_B e^{-u^2-v^2}du\ dv=\left(\int_{-2/\sqrt2}^{2/\sqrt2}e^{-t^2}dt\right)^2=\left(\int_{-\sqrt2}^{\sqrt2}e^{-t^2}dt\right)^2=\Big(\sqrt{\pi}\operatorname{erf}\left({\sqrt2}\right)\Big)^2.\end{align}
Therefore
\begin{align}
\iint_D e^{-x^2-y^2} dx\ dy&=\left(\int_{-3/\sqrt2}^{3/\sqrt2}e^{-t^2}dt\right)^2-\left(\int_{-\sqrt2}^{\sqrt2}e^{-t^2}dt\right)^2
\\&=\Biggl(\sqrt{\pi}\operatorname{erf}\left(\frac{3}{\sqrt2}\right)\Biggr)^2-\Big(\sqrt{\pi}\operatorname{erf}\left({\sqrt2}\right)\Big)^2
\\&=\pi\Biggl(\left(\operatorname{erf}\left({\textstyle\frac{3}{\sqrt2}}^{\vphantom{a^2}}_{\vphantom{a_2}}\right)\right)^2-\big(\operatorname{erf}\left({\sqrt2}\right)\big)^2\Biggr)\approx0.26244.
\end{align}
I don't think you can get the answer in terms of elementary functions.  You at least need the error function $\operatorname{erf}$.
A: Since $x^2+y^2$ is invariant under rotations of the $(x,y)$-plane, rotate $e5^\circ$ so the the bounds $2\le|x|+|y|\le 3$ say that the new coordinates, $u,v,$ each lie in specified sets and $e^{-(x^2+y^2)} = e^{-(u^2+v^2)} = e^{-u^2}\cdot e^{-v^2}.$
To be fully explicity: the bounds will say $u$ is in a specified set, independently of $v,$ and similarly $v$ independently of $u.$ That makes it possible to write
$$
\iint\limits_{u\in A,\,\,v\in B} g(u)h(v) \, d(u,v) = \int_A g(u)\,du \int_B h(v)\,dv.
$$
And $A$ and $B$ will both be the same set.
