## Calculus (3rd Edition)

$$\frac{52}{3}$$
Given $$y=6-x^3,\ \ \ y=x^2-6, \ \ \ y=0$$ First, we find the intersection points \begin{align*} 6-x^3&=x^2-6\\ x^3+x^2-12&=0\\ (x-2)\left(x^{2}+3 x+6\right)&=0 \end{align*} Then $x= 2$ and $6-x^3\geq x^2-6$ for $0\leq x\leq 2$; hence, the area is given by \begin{aligned} \text{Area}&=\int_{0}^{2}\left[\left(6-x^{3}\right)-\left(x^{2}-6\right)\right]d x\\ &=\int_{0}^{2}\left( -x^{3} -x^2+12\right) d x\\ &=\left( \frac{-1}{4}x^{4} -\frac{1}{3}x^3+12x\right) \bigg|_{0}^{2}\\ &=\frac{52}{3} \end{aligned} We use the Mathematica program to plot the region.