Answer
$$\frac{52}{3}$$
Work Step by Step
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.
