Answer
The area of the region is $32/3$.
Work Step by Step
$y=x^2-2$ and $y=2$
1) Draft the graph
The graph is enclosed below. From the graph, we choose integration with respect to $x$ to employ here for calculation of the region area.
2) Find the limits of integration:
We can find the limits of integration by finding points of intersection between the curve and the line.
$$x^2-2=2$$ $$x^2=4$$ $$x=\pm2$$
So, the upper limit is $2$ and the lower one is $-2$.
3) Find the area:
Looking at the draft, we see that the region is bounded above by $y=2$ and below by $y=x^2-2$. So, according to definition, the area of the region is
$$A=\int^2_{-2}[2-(x^2-2)]dx=\int^2_{-2}(4-x^2)dx$$ $$A=\Big(4x-\frac{x^3}{3}\Big)\Big]^2_{-2}$$ $$A=\Big(8-\frac{8}{3}\Big)-\Big(-8+\frac{8}{3}\Big)$$ $$A=2\Big(8-\frac{8}{3}\Big)$$ $$A=\frac{32}{3}$$
