Question

Find the volume of the solid generated by revolving the region bounded above by yequals16 cosine x and below by yequals3 secant x​, negative StartFraction pi Over 3 EndFraction less than or equals x less than or equals StartFraction pi Over 3 EndFraction about the​ x-axis.

Answer 1

`V=(256\pi^2)/(3)+46\pi√(3)`

Explanation:

Let the functions be
`y_1` and
`y_2`

`y_1=3\sec x`

`y_2=16\cos x`

The volume of the solid of revolution generated is given by

`V = \int_(-\pi)/(3)^(\pi)/(3) \pi(y_2^2 - y_1^2) dx`

`V = \int_(-\pi)/(3)^(\pi)/(3) \pi(16\cos x)^2 - (3\sec x)^2) dx`

`V = \pi\int_(-\pi)/(3)^(\pi)/(3) 256\cos^2 x - 9\sec^2x) dx`

Using the trigonometric expansion of
`\cos^2x = (1+\cos 2x)/(2)` and integrating it as well as using the fact that the integral of
`\sec^2x=\tan x`, we have

`V =(256\pi^2)/(3)+46\pi√(3)`