Answer
$\frac{16}{3}$
Work Step by Step
The integration area is a quarter of a radius of $ 2. $ The size of the region is given by:
\[
\begin{aligned}
\iint_{R} z d y d x &=\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}} \sqrt{-x^{2}+4} d y d x \\
&=\int_{0}^{2} \sqrt{-x^{2}+4}(\sqrt{-x^{2}+4}-0) d x \\
&=\int_{0}^{2}\left(4-x^{2}\right) d x \\
&=\frac{16}{3}
\end{aligned}
\]
