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

Answer:

$V=\dfrac{256\pi^2}{3}+46\pi\sqrt{3}$

Step-by-step 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

$V = \int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} \pi(y_2^2 - y_1^2) dx$

$V = \int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} \pi(16\cos x)^2 - (3\sec x)^2) dx$

$V = \pi\int_\dfrac{-\pi}{3}^\dfrac{\pi}{3} 256\cos^2 x - 9\sec^2x) dx$

Using the trigonometric expansion of $\cos^2x = \dfrac{1+\cos 2x}{2}$ and integrating it as well as using the fact that the integral of $\sec^2x=\tan x$, we have

$V =\dfrac{256\pi^2}{3}+46\pi\sqrt{3}$