Integration on 4th dimension unit ball I'm trying to calculate \begin{equation*}\int _B e^{x^2+y^2-z^2-w^2}\end{equation*}where \begin{equation*}B=\{\vec x\in\mathbb R^4:||x||\le 1\}\end{equation*}is the unit ball in 4 dimensions.
I tried using the standard ball coordinates and rewriting the function as \begin{equation*}e^{x^2+y^2-z^2-w^2}=e^{r^2-2(z^2+w^2)}\end{equation*}where $r^2=x^2+y^2+z^2+w^2$. It did eliminate some of the variables but I still don't know how to integrate it.

How about $r:=\sqrt{x^2+y^2}$ and $s:=\sqrt{z^2+w^2}$?  Let $\theta$ and $\phi$ be the usual angular coordinates of the $xy$-plane and the $zw$-plane, respectively.  Then, $\text{d}x\,\text{d}y=r\,\text{d}r\,\text{d}\theta$ and $\text{d}z\,\text{d}w=s\,\text{d}s\,\text{d}\phi$.  The ball $B$ is then given by $r^2+s^2\leq 1$, $0 \leq \theta \leq 2\pi$, and $0\leq \phi \leq 2\pi$.  Hence, the required integral is $\iiiint_B\,rs\,\exp\left(r^2-s^2\right)\,\text{d}r\,\text{d}\theta\,\text{d}s\,\text{d}\phi=(2\pi)^2\,\iint_D\,rs\,\exp\left(r^2-s^2\right)\,\text{d}r\,\text{d}s$, where $D$ is a quarter of a disc $\left\{(r,s)\in\mathbb{R}^2\,|\,r^2+s^2\leq 1\text{ and }r,s\geq 0\right\}$.
You can then again define $\rho:=\sqrt{r^2+s^2}$ and $\omega$ to be the usual angular coordinate in the $rs$-plane.  Hence, $\text{d}r\,\text{d}s=\rho\,\text{d}\rho\,\text{d}\omega$.  Now, $D$ is defined by $0\leq \rho \leq 1$ and $0 \leq \omega \leq \frac{\pi}{2}$.  Your integral becomes $(2\pi)^2\,\iint_D\,\rho^3\cos(\omega)\sin(\omega)\exp\left(\rho^2\cos(2\omega)\right)\,\text{d}\rho\,\text{d}\omega$.  I believe you can finish this.  If my calculation is correct, the answer is $\left(\text{e}+\frac{1}{\text{e}}-2\right)\pi^2$.
