Answer
$$A=\frac{16}{3}$$
Work Step by Step
To find the area of the region, we consider the following integral.
\begin{align}
A&=\int_{x=0}^{x=2} \int_{y=0}^{y=4-x^{2}} d y d x\\
&=\int_{x=0}^{x=2}[y]_{0}^{4-x^{2}} d x\\
&=\int_{x=0}^{x=2}\left(4-x^{2}-0\right) d x \\
&=\left[4 x-\frac{x^{3}}{3}\right]_{0}^{2}\\
&=\left[4(2) -\frac{2^{3}}{3}-0\right] \\
&=\frac{16}{3}
\end{align}
