Answer
$$4\pi $$
Work Step by Step
The area is given by
\begin{align*}
A&=\int_{-2}^{2}\left(y_{1}(x)-y_{2}(x)\right) d x\\
&=2 \int_{-2}^{2} \sqrt{4-x^{2}} d x\\
&=4 \int_{0}^{2} \sqrt{4-x^{2}} d x
\end{align*}
Let $x=2\sin u, \ \ dx = 2\cos udu$. Then
\begin{align*}
A &=4 \int_{0}^{2} \sqrt{4-x^{2}} d x\\
&=4\int_{0}^{\pi/4} 4\cos^2 udu\\
&=8\int_{0}^{\pi/4} [1+\cos2 u]du\\
&=8[u+\frac{1}{2}\sin 2u]\bigg|_{0}^{\pi/2}\\
&=4\pi
\end{align*}
