Answer
True
Work Step by Step
In polar coordinates, $D=\{(r,\theta)|0\leq r\leq 2,0\leq \theta\leq 2\pi\}$.
Converting to polar coordinates,
$\iint_D\sqrt{4-x^2-y^2}dA=\int_0^2\int_0^{2\pi}\sqrt{4-r^2}\cdot rd\theta dr$
$=\int_0^2\int_0^{2\pi}r\sqrt{4-r^2}d\theta dr$
$=\int_0^22\pi r\sqrt{4-r^2}dr$
$=-\frac{2\pi}{3}(4-r^2)^{3/2}]_0^2$
$=0-(-\frac{2\pi}{3}\cdot 8)$
$=\frac{16\pi}{3}$
Then, the statement is true.
