Spherical Coordiantes in vector calc I need to evaluate this triple integral
$$ \int^2_{-2} \int_0^{\sqrt{4-x^2}} \int_0^{\sqrt{4-x^2-y^2}} (x^2+y^2+z^2) \, dz \, dy \, dx $$

My solution:
First I identified the solid as being a quarter of an ice cream cone bounded by cone $z=r$ and sphere $z=\sqrt{4-r^2}$. The line of intersection of the cone and there sphere is found from $z=\sqrt{4-r^2}$, thus $z=r=3$. In the xy-plane we have a quarter of a circle $ 0\leq\theta\leq\pi/2, 0\leq r\leq 3 $. Also, $ r\leq z\leq\sqrt{4-r^2} $. 
The solid in spherical coordinates is $ 0\leq p\leq\sqrt{4}, 0\leq\alpha\leq\pi/4, 0\leq\theta\leq\pi/2 $. Then the integrand is $p^2$
Just wondering if this is correct and if it is how to set up the triple integration from here and then evaluate.
 A: Let $E$ be the solid in question.  The limits of the $z$ integral are $z=0$ and $z=\sqrt{4-x^2-y^2}$.  That means the bottom face of the solid is in the $xy$-plane and the top is in the sphere $x^2 + y^2 + z^2=4$.  
Looking at the $x$ and $y$ limits, we see that the projection of $E$ onto the $xy$-plane is the region
$$
    D = \left\{(x,y)\mid -2 \leq x \leq 2,\ 0 \leq y \leq \sqrt{4-x^2}\right\}
$$
This is the upper half disk of radius $2$ centered at the origin.
So $E$ is the part of the ball of radius $2$ that satisfies $y \geq 0$ and $z\geq 0$.  
Draw this solid to visualize it in spherical coordinates.  We have $0 \leq \theta\leq\pi$ (half of the circle in the equatorial plane), $0 \leq \phi\leq \frac{\pi}{2}$ (from the north pole to the equator), and $0 \leq \rho\leq 2$ (center to the surface).
So
$$\begin{split}
\iiint_E (x^2+y^2+z^2)\,dV
&= \int_0^\pi \int_0^{\pi/2} \int_0^2\rho^2(\rho^2\sin\phi)\,d\rho\,d\phi\,d\theta\\
&= \int_0^\pi \int_0^{\pi/2} \int_0^2\rho^4\sin\phi\,d\rho\,d\phi\,d\theta\\
&= \int_0^\pi 1\,d\theta \cdot \int_0^{\pi/2} \sin\phi\,d\phi \cdot \int_0^2\rho^4\,d\rho \\
&= \pi \cdot 1 \cdot \frac{32}{5} = \frac{32\pi}{5}
\end{split}$$
A: Let $r^2=x^2+y^2+z^2$. By Cavalieri's principle/the shell method the integral of $r^2$ over the set $r\leq 2$ is given by
$$\int_{0}^{2} \rho^2\cdot 4\pi\rho^2\,d\rho = \frac{128\pi}{5}$$
and by symmetry your integral is just one fourth of the previous quantity.
