Count a triple integral on a sphere $\int\int\int_{x^2+y^2+z^2 \le 1} |x+y|e^{2xy-z^2}dxdydz$
I have tried to change coordinates to spherical, but it didn't help. Can you give me a hint, please?

Use the substitution 
$$\begin{cases} u = \frac{x+y}{\sqrt{2}} \\ v = \frac{x-y}{\sqrt{2}} \\ w = z\\ \end{cases} \implies u^2-v^2 = 2xy$$
(chosen so that the Jacobian is $1$) to get the following integral
$$\sqrt{2}\iiint_{u^2+v^2+w^2\leq 1} |u|e^{u^2-v^2-w^2}\:du \:dv \:dw = \sqrt{2} \iint_{v^2+w^2\leq 1} e^{1-2(v^2+w^2)}\: dv \:dw$$
by using symmetry (the integrand is an even function of $u$, so double the value and then integrate from $u=0$ to $u=\sqrt{1-v^2-w^2}$). Then convert to "polar":
$$ = 2\sqrt{2}\pi e \int_0^1 r\:e^{-2r^2}dr = \frac{\pi(e^2-1)}{e\sqrt{2}}$$

$\textbf{EDIT}$: Alternatively, at the $uvw$ stage, convert to spherical coordinates with $u$ as the "$z$". The integrand becomes
$$ r^3|\cos\theta|\sin\theta e^{r^2(\cos^2\theta-\sin^2\theta)} = \frac{1}{2}r^3\sin2\theta e^{r^2\cos2\theta}$$
so it becomes clear which order of integration is easiest. Using symmetry, we only have to integrate the "upper" half to avoid the absolute value.
$$= 2\sqrt{2}\int_0^{2\pi} \int_0^1 \int_0^{\frac{\pi}{2}}\frac{1}{2}r^3\sin2\theta e^{r^2\cos2\theta} \:d\theta \:dr \:d\phi = \pi\sqrt{2}\int_0^1 -re^{r^2\cos2\theta}\Biggr|_0^{\frac{\pi}{2}}dr$$ 
$$=\pi\sqrt{2}\int_0^1 2r\sinh(r^2)dr = \pi \sqrt{2}\cosh(r^2)\Biggr|_0^1 = \pi\sqrt{2}(\cosh(1)-1)$$
which is the same thing.
