$\iiint_R (x^2+y^2+z^2)^\frac{3}{2}e^{-x^2-y^2-z^2}dV_{xyz}$ im having trouble evaluating this triple integral
$\iiint_R (x^2+y^2+z^2)^\frac{3}{2}e^{-x^2-y^2-z^2}dV_{xyz}$
$R = (x,y,z):1\le x^2+y^2+z^2\le9$
any help is appreciated as always.
 A: Hint:
Whenever you see $x^2 + y^2 + z^2$, this is begging to be transformed to $\rho^2 = x^2 + y^2 + z^2$ (spherical coordinates).
So, applying that substitution (and including the Jacobian):
$$\iiint_G (\rho^3e^{-\rho^2})\rho^2 \sin(\varphi)dV_{\rho,\varphi,\theta}$$
EDIT: For converting the region (as seems to be your questions from the comments), we can see that it's a sphere of radius $3$ with a hole of radius $1$ (as Ron Gordon noted).  To model this, we let $\rho$ range from $1$ to $3$, $\varphi$ from $0$ to $\pi$ and $\theta$ from $0$ to $2\pi$.
A: Let's consider a different coordinate system, a spherical one to be precise, noting $x^2+y^2+z^2=r^2$. This means we then express our domain in spherical coordinates as
$$R=\{(r,\theta,\phi)|1\le r^2\le9,0\le\theta\lt\pi,0\le\phi\lt2\pi\}$$
i.e. a closed ball of radius 3 with a spherical hole of radius 1 at its core, centered at the origin. 
Recalling that $\mathrm{d}V=r^2\sin\theta\,\mathrm{d}r\,\mathrm{d}\theta\,\mathrm{d}\phi$, we may rewrite our integral as an iterated integral in spherical coordinates:
$$\begin{align*}\iiint\limits_{R}(x^2+y^2+z^2)^\frac32e^{-(x^2+y^2+z^2)}\mathrm{d}V&=\iiint\limits_{R}r^5e^{-r^2}\mathrm{d}V\\&=\int_0^{2\pi}\mathrm{d}\phi\int_0^\pi\mathrm{d}\theta\sin\theta\int_1^3\mathrm{d}r\,r^5e^{-r^2}\end{align*}$$ 
