Solve the triple integral $\iiint_D (x^2 + y^2 + z^2)\, dxdydz$ How does one go about solving the integral: 
$$
\iiint_D (x^2 + y^2 + z^2)\, dxdydz, 
$$
where 
$$
D=\{(x,y,z) \in \mathbb{R}^3: x^2 + y^2 + z^2 \le 9\}. 
$$
I believe I am supposed to convert to spherical coordinates but I would need some help with how this is done and what the answer to this integral would be.
Thanks in advance!
 A: The easiest way to do this is to make a switch to spherical coordinates. There $\rho^2=x^2+y^2+z^2$ and $dxdydz=\rho^2 \sin \phi \,\,d\rho d\theta d\varphi$. So we have $$\iiint_Dx^2+y^2+z^2 dxdydz=\iiint_D \rho^2 \cdot \rho^2 \sin \varphi\,d\rho\theta d\varphi$$ Now we are integrating over a region $D$. What is $D$? It is a sphere of radius $3$ centered at the origin. So $0\leq \rho \leq 3$, $0\leq \theta\leq 2\pi$, and $0\leq \varphi \leq \pi$. 
$$\iiint_D \rho^4\sin \varphi \, d\rho d \theta d\varphi=\int_{0}^{\pi}\int_{0}^{2\pi}\int_{0}^3 \rho^4 \sin \varphi d\rho d\theta d\varphi$$
Which is more easily integrated as $$\int_{0}^{2\pi}d\theta \int_{0}^\pi \sin \varphi \,d \varphi\int_0^3 \rho^4 d\rho=\frac{972\pi}{5}$$
Edit: I must have had a stroke. The answer has been corrected.
A: A quick way to evaluate it is to note that the volume of the spherical shell from radius $r$ to radius $r + \Delta r$ is approximately $4\pi r^2 \Delta r$, so your result should be
$$\int_0^3 r^2 (4\pi r^2) \,dr$$
$$= {4 \over 5} \pi r^5\bigg|_{r=0}^3$$
$$= {4 \over 5} 3^5 \pi$$
$$={972 \pi \over 5}$$
To do it properly you should do spherical coordinates like mathematics2x2life is trying to do.
A: $$
\int_{0}^{3}{\rm d}r\,r^{2}\times r^{2}\quad
\overbrace{\int{\rm d}\Omega_{\,\vec{r}}}^{4\pi}\
=\
4\pi\,{3^{5} \over 5}
=
{972 \over 5}\,\pi
$$
