Answer
The statement is true.
Work Step by Step
$D$ is the disk given by $x^{2}+y^{2}\leq 4$; then
$$
V=\iint_{D} \sqrt{4-x^{2}-y^{2}} d A
$$
is the volume under the surface $ x^{2}+y^{2}+z^{2}=4$ and above the $x y$-plane which is equal to half of the volume of the sphere $ x^{2}+y^{2}+z^{2}=4$
$$
V=\frac{1}{2} \cdot \frac{4}{3} \pi(2)^{3}=\frac{16}{3} \pi
$$
Therefore,
$$
V=\iint_{D} \sqrt{4-x^{2}-y^{2}} d A =\frac{16}{3} \pi.
$$
So, the statement is true.
