## Calculus (3rd Edition)

$$\frac{64}{3}$$
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.