Answer
$$\frac{64}{3}$$
Work Step by Step
Given $$x=y^{2}-5 ,\ \ \ x =3-y^{2} $$
First we find the intersection points
\begin{aligned}
y^{2}-5 &=3-y^{2} \\
2 y^{2}-8 &=0
\end{aligned}
Then $ y=\pm 2$ and $ 3-y^{2}\geq y^{2} -5 $ for $-2\leq y\leq 2$; hence, the area is given by
\begin{aligned}
\text{Area}&=\int_{-2}^{2}\left(3-y^{2}\right)-\left(y^{2}-5\right) d y\\
&=\int_{-2}^{2}\left(8-2 y^{2}\right) d y\\
&=2\int_{0}^{2}\left(8-2 y^{2}\right) d y\\
&=2 \left(8 y-\frac{2}{3} y^{3}\right)\bigg|_{0} ^{2}\\
&=\frac{64}{3}
\end{aligned}
We use the Mathematica program to plot the area.
