the problem is: $$I=\int_{B^4} e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw $$ where $B^4$ is the unit 4 ball, explicitly: $$ B^4=\{(x,y,z,w)\in\Bbb{R}^4 \mid x^2+y^2+z^2+w^2\leq 1\}$$ here is my solution:
using fubini's theorem we obtain: $$I=\int_{B^2}e^{-z^2-w^2}\left(\int_{{B^2}_{\sqrt{1-z^2-w^2}}} e^{x^2+y^2} \, dx\, dy\right)\,dz\,dw$$ where $B^2$ is the unit 2 ball and ${B^2}_{\sqrt{1-z^2-w^2}}$ is the 2 ball centred at the origin with radius $\sqrt{1-z^2-w^2}$.
next we calculate $$\int_{{B^2}_{\sqrt{1-z^2-w^2}}}e^{x^2+y^2}dxdy$$ using polar cordinates, so: $$\int_{{B^2}_{\sqrt{1-z^2-w^2}}}e^{x^2+y^2}dxdy=\int_{0}^{2\pi}\int_{0}^{\sqrt{1-z^2-w^2}}e^{r^2}rdrd\theta=\pi(e^{1-z^2-w^2}-1)$$ plugging this result into the outer integral we get: $$I=\int_{B^2}e^{-z^2-w^2}(\pi(e^{1-z^2-w^2}-1))dzdw=\pi e \int_{B^2}e^{-2(z^2+w^2)}dzdw-\pi \int_{B^2}e^{-z^2-w^2}dzdw$$ again using polar cordinates we receive: $$\int_{B^2}e^{-2(z^2+w^2)}dzdw=\int_{0}^{2\pi}\int_{0}^{1}e^{-2r^2}rdrd\theta=\frac{\pi}{2}(1-e^{-2})$$ and: $$\int_{B^2}e^{-z^2-w^2}dzdw=\int_0^{2\pi}\int_0^1re^{-r^2}drd\theta=\pi(1-\frac{1}{e})$$ plugging this in we get: $$I=\pi^2(\sinh(1)+1-\frac{1}{e})$$ is this solution correct or have I made a mistake along the way?
could you verify using two sets of polar coordinates e.g.: $$r_1=\sqrt{x^2+y^2}$$ $$r_2=\sqrt{z^2+w^2}$$ $$x=r_1\cos\theta_1,y=r_1\sin\theta_1,z=r_2\cos\theta_2,w=r_2\sin\theta_2$$ $$dxdydzdx=r_1r_2dr_1dr_2d\theta_1d\theta_2$$ and we know that $r_1^2+r_2^2\le1$ and our integral would become: $$\iiiint_{\Omega^4}e^{r_1^2-r_2^2}r_1r_2dr_1d\theta_1dr_2d\theta_2$$ Then we would just need to find a condition for $\theta_1,\theta_2$