<s>[INST] verifying the result $\int_{B^4}e^{x^2+y^2-z^2-w^2}\,dx\,dy\,dz\,dw=\pi^2(\sinh(1)+1-\frac{1}{e})$ 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?
 [/INST] 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 $\theta_1,\theta_2$
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