Answer:
Step-by-step explanation:
The curves and the region bounded are shown in the attachment.
The region is revolved around x axis.
We have to find the volume of the resulting solid
Limits for x are [tex]\frac{-\pi}{3} ,\frac{\pi}{3}[/tex]
The region lies above 3 secx =y and below y= 12 cos x
So volume= [tex]\pi\int\limits_ {\frac{-\pi}{3}} ^ {\frac{\pi}{3}} \, (12cosx)^2 - (3 secx)^2 dxdx \\= \pi \int\limits_ {\frac{-\pi}{3}} ^ {\frac{\pi}{3}} (144 cos^2 x dx -9 sec^2 x dx)\\= \pi \int\limits_ {\frac{-\pi}{3}} ^ {\frac{\pi}{3}}(72 +72 cos2x -9sec^2 x )dx\\= 72 +36 sin 2x -9 tanx[/tex]
substitute limits
=[tex]6\pi (3\sqrt{3} +8\pi)[/tex] cubic units
