Solve the problem.
-Find the volume of the solid generated by revolving the region bounded by the curve $y = 4 \cos x$ and the $x - a x i s$, $\frac { \pi } { 2 } \leq x \leq \frac { 3 \pi } { 2 }$, about the $x$-axis.
A) $8 \pi ^ { 2 }$
B) $16 \pi$
C) $16 \pi ^ { 2 }$
D) $8 \pi ^ { 3 }$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region around the x-axis is calculated using the disk method: $ V = \int_{a}^{b} \pi [f(x)]^2 \, dx $. Here, $ f(x) = 4 \cos x $ and the limits are $ \frac{\pi}{2} $ to $ \frac{3\pi}{2} $. The integral becomes $ V = \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \pi (4 \cos x)^2 \, dx = 16\pi \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} \cos^2 x \, dx $. Using the identity $ \cos^2 x = \frac{1 + \cos 2x}{2} $, the integral simplifies to $ 8\pi \int_{\frac{\pi}{2}}^{\frac{3\pi}{2}} (1 + \cos 2x) \, dx $. Evaluating this gives $ 8\pi [x + \frac{1}{2} \sin 2x]_{\frac{\pi}{2}}^{\frac{3\pi}{2}} = 8\pi [\pi] = 8\pi^2 $.

Solve the Problem y=4cos⁡xy = 4 \cos xy=4cosx And The x−axisx - a x i sx−axis

Solve the Problem $y = 4 \cos x$ And The $x - a x i s$