Answer:
[tex]V=\dfrac{256\pi^2}{3}+46\pi\sqrt{3}[/tex]
Step-by-step explanation:
Let the functions be [tex]y_1[/tex] and [tex]y_2[/tex]
[tex]y_1=3\sec x[/tex]
[tex]y_2=16\cos x[/tex]
The volume of the solid of revolution generated is given by
[tex]V = \int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} \pi(y_2^2 - y_1^2) dx[/tex]
[tex]V = \int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} \pi(16\cos x)^2 - (3\sec x)^2) dx[/tex]
[tex]V = \pi\int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} 256\cos^2 x - 9\sec^2x) dx[/tex]
Using the trigonometric expansion of [tex]\cos^2x = \dfrac{1+\cos 2x}{2}[/tex] and integrating it as well as using the fact that the integral of [tex]\sec^2x=\tan x[/tex], we have
[tex]V =\dfrac{256\pi^2}{3}+46\pi\sqrt{3}[/tex]
