Answer
The area of the region is $6\sqrt3$.
Work Step by Step
$y=8\cos x$ and $y=\sec^2x$ for $-\pi/3\le x\le\pi/3$
1) Draft the graph:
The graph is drafted below. We would use integration with respect to $x$ in this exercise.
2) Find limits of integration:
To find the limits of integration, we will find the intersection points.
$$8\cos x-\sec^2x=0$$ $$8\cos x-\frac{1}{\cos^2x}=0$$ $$\frac{8\cos^3x-1}{\cos^2x}=0$$ $$8\cos^3x-1=0$$ $$\cos x=\frac{1}{2}$$ $$x=\pm\frac{\pi}{3}$$
Therefore, the upper limit is $\pi/3$ and the lower one is $-\pi/3$.
3) Find the area:
The region is bounded above by $y=8\cos x$ and below by $y=\sec^2x$. So according to definition, the area of the region is $$A=\int^{\pi/3}_{-\pi/3}(8\cos x-\sec^2x)dx$$ $$A=8\sin x-\tan x\Big]^{\pi/3}_{-\pi/3}$$ $$A=\Big(\frac{8\sqrt3}{2}-\sqrt3\Big)+\Big(\frac{8\sqrt3}{2}-\sqrt3\Big)$$ $$A=2\Big(\frac{8\sqrt3}{2}-\sqrt3\Big)$$ $$A=2\Big(\frac{8\sqrt3-2\sqrt3}{2}\Big)=6\sqrt3$$
