TITLE: Calculate the integral ...
QUESTION [6 upvotes]: We need to find the integral
$$\iiiint\limits_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}dxdydudv$$
I was only able to get to this point ...
$$\iiiint\limits_{x^2+y^2+u^2+v^2\leq 1}e^{x^2+y^2-u^2-v^2}dxdydudv=\iint\limits_{x^2+y^2\leq 1}e^{x^2+y^2}\left ( \iint\limits_{u^2+v^2\leq 1-x^2-y^2}\frac{dudv}{e^{u^2+v^2}} \right )dxdy$$
I don’t know how to solve it further ...

REPLY [3 votes]: Use polar coordinates (twice):
In the inner integral change variables
$u\rightarrow r\cos{\theta}$,
$v\rightarrow r\sin{\theta}$, to get
$$\iint\limits_{u^2+v^2\leq 1-x^2-y^2}\frac{dudv}{e^{u^2+v^2}}=
\int_{0}^{2\pi}\int_{0}^{\sqrt{1-x^2-y^2}}r e^{-r^2} dr\,d\theta
=\pi \left(1-e^{-1}e^{x^2+y^2}\right)$$
Then use polar coordinates once more to compute
$$\int_{0\leq x^2+y^2\leq 1} e^{x^2+y^2} dx dy=\int_{0}^{2\pi} \int_{0}^{1}
r e^{r^2} dr d\theta=\pi (e-1)$$