Answer
The solid is inside both the cylinder $x^{2}+y^{2} =4$ and the ellipsoid $4x^{2}+4y^{2} + z^{2} = 64 $
The required volume is given by:
$$
\begin{aligned}
V &=\iint_{x^{2}+y^{2} \leq 4}[\sqrt{64-4 x^{2}-4 y^{2}}-(-\sqrt{64-4 x^{2}-4 y^{2}})] d A \\
&=\frac{8 \pi}{3}(64-24 \sqrt{3})
\end{aligned}
$$
Work Step by Step
The solid is inside both the cylinder $x^{2}+y^{2} =4$ and the ellipsoid $4x^{2}+4y^{2} + z^{2} = 64 $,
i.e. between the surfaces $z = \sqrt { 64 -4x^{2}+4y^{2}} $ and $z = -\sqrt { 64 -4x^{2}+4y^{2}}. $ So the volume is
$$
\begin{aligned}
V& =\iint_{D}(Z_{1}-Z_{2}) d A\\
&=\iint_{x^{2}+y^{2} \leq 4}\left(\sqrt { 64 -4x^{2}+4y^{2}} -(-\sqrt { 64 -4x^{2}+4y^{2}}) \right) d A\\
\end{aligned}
$$
In polar coordinates, we have $x^{2}+y^{2}=r^{2} $; thus the disk D is given by:
$$ D=\left\{(r, \theta) | 0 \leq r \leq 2, \quad 0 \leq \theta \leq 2\pi \right\} $$
and, the required volume is given by:
$$
\begin{aligned}
V &=\iint_{x^{2}+y^{2} \leq 4}[\sqrt{64-4 x^{2}-4 y^{2}}-(-\sqrt{64-4 x^{2}-4 y^{2}})] d A \\
&=\iint_{x^{2}+y^{2} \leq 4} 2 \sqrt{64-4 x^{2}-4 y^{2}} d A \\
&=4 \int_{0}^{2 \pi} \int_{0}^{2} \sqrt{16-r^{2}} r d r d \theta \\
&=4 \int_{0}^{2 \pi} d \theta \int_{0}^{2} r \sqrt{16-r^{2}} d r \\
&=4[\theta]_{0}^{2 \pi}\left[-\frac{1}{3}\left(16-r^{2}\right)^{3 / 2}\right]_{0}^{2} \\
&=8 \pi\left(-\frac{1}{3}\right)\left(12^{3 / 2}-16^{2 / 3}\right) \\
&=\frac{8 \pi}{3}(64-24 \sqrt{3})
\end{aligned}
$$
