Answer
The solid is inside the sphere $x^{2}+y^{2} +z^{2}= 16$ and outside the cylinder $x^{2}+y^{2} = 4$.
The volume is
$$
\begin{aligned}
V& = \iint_{4 \leq x^{2}+y^{2} \leq 16} z d A\\
&=32 \sqrt{3} \pi
\end{aligned}
$$
Work Step by Step
The solid is inside the sphere $x^{2}+y^{2} +z^{2}= 16$ and outside the cylinder $x^{2}+y^{2} = 4$.
If we put $z = 0$ in the equation of the sphere, we get $x^{2}+y^{2} = 16$. This means that the sphere intersects the $xy-$plane in the circle $x^{2}+y^{2} = 4$ ,
so the volume is
$$
\begin{aligned}
V& = \iint_{4 \leq x^{2}+y^{2} \leq 16} z d A
& \quad\left[\begin{array}{c}{ \text {by symmetry } } \end{array}\right] \\
& =2 \iint_{4 \leq x^{2}+y^{2} \leq 16} \sqrt{16-x^{2}-y^{2}} d A.
\end{aligned}
$$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $. Thus, the disk D is given by:
$$ D=\left\{(r, \theta) | 2 \leq r \leq 4, \quad 0 \leq \theta \leq 2\pi \right\} $$
and, by Formula 3, we have
$$
\begin{aligned}
V& = \iint_{4 \leq x^{2}+y^{2} \leq 16} z d A\\
& =2 \iint_{4 \leq x^{2}+y^{2} \leq 16} \sqrt{16-x^{2}-y^{2}} d A\\
& =2 \int_{0}^{2 \pi} \int_{2}^{4} \sqrt{16-r^{2}} r d r d \theta \\
&=2 \int_{0}^{2 \pi} d \theta \int_{2}^{4} r\left(16-r^{2}\right)^{1 / 2} d r\\
& =2[\theta]_{0}^{2 \pi}\left[-\frac{1}{3}\left(16-r^{2}\right)^{3 / 2}\right]_{2}^{4} \\
&=-\frac{2}{3}(2 \pi)\left(0-12^{3 / 2}\right) \\
&=\frac{4 \pi}{3}(12 \sqrt{12}) \\
&=32 \sqrt{3} \pi.
\end{aligned}
$$
