## Calculus (3rd Edition)

$$\frac{64}{3}$$
Given $$y=4-x^2,\ \ \ y=x^2-4$$ First, we find the intersection points \begin{align*} 4-x^2&=x^2-4\\ 2x^2-8&=0 \end{align*} Then $x=\pm 2$ and $4-x^2\geq x^2-4$ for $-2\leq x\leq 2$; hence, the area is given by \begin{aligned} \text{Area}&=\int_{-2}^{2}\left[\left(4-x^{2}\right)-\left(x^{2}-4\right)\right]d x\\ &=\int_{-2}^{2}\left(8-2 x^{2}\right) d x\\ &=2 \int_{0}^{2}\left(8-2 x^{2}\right) d x\\ &=2\left(8 x-\frac{2}{3} x^{3}\right)\bigg|_{0}^{2}\\ &=\frac{64}{3} \end{aligned} We use the Mathematica program to plot the region.