Find the volume of the solid generated by revolving the region about the given line.
-The region in the first quadrant bounded above by the line $y = 1$, below by the curve $y = \sqrt { \sin 2 x }$, and on the left by the $y$-axis, about the line $y = 1$
A) $\frac { \pi ^ { 2 } } { 4 } - \frac { \pi } { 2 }$
B) $\frac { \pi ^ { 2 } } { 4 } + \frac { \pi } { 2 }$
C) $\frac { \pi } { 2 } + 2$
D) $\frac { \pi } { 4 } - \frac { 1 } { 2 }$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region about the line $y = 1$ can be found using the method of washers or disks. The volume $V$ is given by the integral $$V = \pi \int_{a}^{b} (R^2 - r^2) \, dx$$ where $R$ is the outer radius from the axis of rotation to the outer curve and $r$ is the inner radius from the axis of rotation to the inner curve. Here, $R = 1$ and $r = 1 - \sqrt{\sin 2x}$. The region is bounded on the left by the $y$-axis ($x=0$) and extends to where $\sin 2x = 1$, which is at $x = \frac{\pi}{4}$. Thus, the integral becomes $$V = \pi \int_{0}^{\frac{\pi}{4}} \left(1^2 - (1 - \sqrt{\sin 2x})^2\right) \, dx$$. Simplifying and solving this integral gives the volume of the solid, which matches the expression in choice A: $\frac{\pi^2}{4} - \frac{\pi}{2}$.

Find the Volume of the Solid Generated by Revolving the Region