How to determine the value of the definite integral $\iint\limits_H\exp(-(x+y)^2)dxdy$, where $H$ is the area where $x, y \geq 0$. Suppose I have a function$f(x,y) = \exp(-(x+y)^2)$ and the area $H\subset \mathbb{R}^2$, where I wish to know $$\iint\limits_H\exp(-(x+y)^2)dxdy.$$
How does one solve this?
I had an idea based on how $f$ looks, which tries to transform the entire expression to a circle by writing it in the form of $\exp(i\pi(...))$, but I couldn't work out the details.
So then I thought I might be able to split it up and calculate
$$\int_0^\infty \int_0^\infty\exp(-(x+y)^2)dxdy,$$
but throwing that in to wolframalpha I get to deal with the error function, which I barely know anything about.
 A: Split the first quadrant $H$ up into infinitesimal trapezoids
$$s\leq x+y\leq s+ds$$
of area ${1\over2}\bigl(s+ds)^2-s^2\bigr)\approx s\> ds$ and obtain
$$\int_H\exp\bigl(-(x+y)^2\bigr)\>{\rm d}(x,y)=\int_0^\infty e^{-s^2}\>s\>ds=-{1\over2}e^{-s^2}\biggr|_0^\infty={1\over2}\ .$$
For a more formal treatment parametrize $H$ by
$$x=s-t,\quad y=s+t\qquad(0\leq|t|\leq s)\ .$$
The Jacobian is $\equiv2$. Therefore we obtain
$$\int_H \exp\bigl(-(x+y)^2\bigr)\>{\rm d}(x,y)=2\int_0^\infty\int_{-s}^s e^{-4s^2}\>dt\>ds=2\int_0^\infty 2s e^{-4s^2}\>ds={1\over2}\ .$$
A: Based on a comment, I have editted my post as follows: 
Let us put $x = r \cos \theta$ and $y = r \sin \theta$, where $0 \leq r < +\infty$ and $0 \leq \theta < \pi / 2$. Then we calculate the Jacobian determinant as follows: 
$$ \frac{\partial(x,y)}{\partial(r, \theta) } = \left| \matrix{ \frac{\partial x}{ \partial r} \  & \   \frac{\partial y }{ \partial r} \\  \frac{\partial x}{ \partial \theta } \ & \  \frac{\partial y }{ \partial \theta } } \right| = \left|  \matrix{ \cos \theta \ & \ \sin \theta  \\ -r \sin \theta \ & \ r \cos \theta } \right| = r. $$
So 
$$ \iint_H \exp{ \left( - (x+y)^2 \right) } \ \mathrm{d}x \mathrm{d} y = \int_0^{ \pi / 2} \int_0^{+\infty}  \exp{ \left( - r^2 (1 + \sin 2 \theta ) \right) }  r  \ \mathrm{d} r \mathrm{d} \theta = \ldots. $$
Hope this will help. 
