Integral in spherical coordinates Let Ω be the solid bounded below by the half-cone $z = \sqrt{x^2 + y^2}$ and above by the spherical surface $x^2 + y^2 + z^2 = 1$. Use spherical coordinates to evaluate the integral 
$\iiint_\Omega e^{(x^2+y^2+z^2)^{3/2}}$ $dxdydz$
Ive been bashing against this, but to no avail, any assistance would be greatly appreciated

Spherical coordinates:
$$
x = r \sin\theta \cos\varphi\\
y = r \sin\theta \sin\varphi\\
z = r \cos\theta.
$$
The volume element:
$$
dxdydz = dr\, rd\theta\, r\sin\theta d\varphi.
$$
Since there is axial symmetry around the zeta axis, the domain of integration is independent of $\varphi$, the $\theta$ coordinate is restricted by the angle of the cone, whose half opening is $\pi/4$, and finally $r$ ranges from $0$ to $1$ since $\Omega$ is enclosed in the unit sphere; therefore:
$$
\iiint_\Omega e^{(x^2+y^2+z^2)^{3/2}}dx dy dz = 
\int_0^{2\pi} d\varphi \int_{0}^{\pi/4} \sin\theta d\theta \int_0^1 dr e^{r^3}r^2\\
=2\pi \left( 1- \frac{\sqrt 2}{2}\right)\frac{1}{3}\left(e-1\right).
$$
