Let V be the volume of the solid generated by revolving the region bounded above by $y = \sqrt { \cos x }$ and bounded below by the x-axis from x = 0 to $x = \frac { \pi } { 2 };$ about the x-axis. Then V is

A) $\pi$
B) $\frac { 3 \pi } { 2 }$
C) $2 \pi$
D) $\frac { 5 \pi } { 2 }$
E) $3 \pi$

Question

Let V be the volume of the solid generated by revolving the region bounded above by

y = \sqrt { \cos x }

and bounded below by the x-axis from x = 0 to

x = \frac { \pi } { 2 };

about the x-axis. Then V is

A) $\pi$
B) $\frac { 3 \pi } { 2 }$
C) $2 \pi$
D) $\frac { 5 \pi } { 2 }$
E) $3 \pi$

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region bounded by $y = \sqrt{\cos x}$ and the x-axis from $x = 0$ to $x = \frac{\pi}{2}$ about the x-axis can be found using the disk method. The formula for the volume is $\int_{a}^{b} \pi [f(x)]^2 dx$, where $f(x) = \sqrt{\cos x}$. Thus, the integral becomes $\int_{0}^{\frac{\pi}{2}} \pi (\cos x) dx$, which evaluates to $\pi [ \sin x ]_{0}^{\frac{\pi}{2}} = \pi [1 - 0] = \pi$.

Let V Be the Volume of the Solid Generated by Revolving