<s>[INST] Calculate $\int_A e^{x^2+y^2-z^2-w^2} \, dx\,dy\,dz\,dw$ Calculate $\int_A e^{(x^2+y^2-z^2-w^2)}\,dx\,dy\,dz\,dw$ where $A=\{x, y, z, w \in \mathbb{R} \mid x^2+y^2+z^2+w^2\leq1\}$
attempt:
$$\int_A e^{x^2+y^2-z^2-w^2} \,dx\,dy\,dz\,dw
  = \int_0^1 \int_{\mathbb{S}^3_r} e^{x^2+y^2-z^2-w^2} \;\mathrm{d}S \;\mathrm{d}r
  \\= \int_0^1 \int_{\mathbb{S}^3_r} e^{2x^2+2y^2-r} \;\mathrm{d}S \;\mathrm{d}r
  = \int_0^1 \int_0^{2\pi} \int_0^{\pi} e^{2r^2 \sin^2\phi - r} r^2 \sin\phi \;\mathrm{d}\phi \;\mathrm{d}\theta \;\mathrm{d}r
$$
I got stuck here. Any help please?
 [/INST] Note that $$A=\{x^2+y^2\leq1, z^2+w^2\leq1-(x^2+y^2)\}$$
So the integral becomes
$$I=\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2}\int_{\{z^2+w^2\leq1-(x^2+y^2)\}}e^{-(z^2+w^2)}$$
Where the inner integral is by Fubini again:
$$(*) = \int_{-\pi}^{\pi}\int_0^cre^{-r^2}drd\theta$$
where $c=\sqrt{1-(x^2+y^2)}$.
Now, this equals $\pi(1-e^{-c^2})$, so we got
$$I =  \pi\int_{\{x^2+y^2\leq1\}}e^{x^2+y^2} \cdot(1-e^{-(1-(x^2+y^2))})$$
Where from here, it is easier to continue by polar coordinates and similar integral as in $(*)$
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