How to solve this problem using spherical coordinates system? The question is very simple:
Volume inside the solid limited by:$ (X^2+Y^2+Z^2=16), (X^2+Y^2=4)$ 
using SPHERICAL coordinates system.
The final answer however can be checked making a "cylindrical integration".
 A: 
In general, when you integrate a volume that includes the origin, $\theta $ and $\phi $ are integrated between 0and $\pi$ and 0 and $2\pi$, respectively. If the volume is convex (or, more generally, star-shaped as it is called) then the shape is determined by the surface function $r(\theta,\phi)$. For example, in the case of a sphere of radius 4 it is $r(\theta,\phi)=4$, in the case of an spheroid is, for example, $r(\theta,\phi)=(1+0.5\cos\theta)$, in the case of a cube is very complicated, so dont use spherical coordinated to calculate the volume of a cube, but in principle there is no problem.
In this case, as you point out, the volume is a cylinder with height 4 and sphere pieces on the top and bottom. N the direction of these pieces,the surface is just the sphere, and in the other directions the surface of the cylinder fullfils $\sqrt{x^2+y^2}/r(\theta,\phi)=\sin\theta$. The equations below just summarizes all this
$\int_0^{2\pi}d\phi\int_{0}^{\pi}\sin\theta d\theta \int_0^{r(\theta)}r^2 d r$
where 
$r(\theta)=
\begin{cases}
4 & \theta<\pi/6~{\rm or}~\theta>5\pi/6\\
2/\sin\theta & \neg
\end{cases}$
We can integrate in the following way
$\begin{align}
\int_0^{2\pi}d\phi\int_{0}^{\pi}\sin\theta d\theta \int_0^{r(\theta)}r^2 d r\\ &= \int_0^{2\pi}d\phi\int_{0}^{\pi}\sin\theta d\theta \frac{r(\theta)^3}{3}\\
&= 2\pi\int_{0}^{\pi}\sin\theta d\theta \frac{r(\theta)^3}{3}\\
&=2\pi\left(\int_0^{\pi/6}\frac{r(\theta)^3}{3} \sin\theta d\theta +\int_{5\pi/6}^{\pi}\frac{r(\theta)^3}{3} \sin\theta d\theta +
\int_{\pi/6}^{5\pi/6}\frac{r(\theta)^3}{3} \sin\theta d\theta\right)\\
&=2\pi\left(\int_0^{\pi/6}\frac{64}{3} \sin\theta d\theta +\int_{5pi/6}^{\pi}\frac{64}{3} \sin\theta d\theta +
\int_{\pi/6}^{5\pi/6}\frac{8}{3\sin^3\theta} \sin\theta d\theta\right)\\
&=2\pi \left(64\frac{(2-\sqrt{3})}{3}+\int_{\pi/6}^{5\pi/6}\frac{8\csc^2\theta}{3} d\theta\right)\\
&=2\pi \left(64\frac{(2-\sqrt{3})}{3}+\frac{16}{\sqrt{3}} \right)\\
&\sim 93.96
\end{align}
$
Note that the integrals are symmetric respecto to $\theta=\pi/2$.
