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 $y = \sqrt { \sin 4 x }$, and on the left by the $y$-axis, about the line $y = 1$
A) $\frac { \pi ^ { 2 } } { 8 } + \frac { \pi } { 4 }$
B) $\frac { \pi } { 8 } - \frac { 1 } { 4 }$
C) $\frac { \pi } { 4 } + 4$
D) $\frac { \pi ^ { 2 } } { 8 } - \frac { \pi } { 4 }$

Question

Quizplus · Accepted Answer

The volume of the solid generated by revolving the region around the line $y = 1$ can be found using the method of washers or disks, integrating with respect to $x$. The volume $V$ is given by the integral of the area of each circular disk, which is $\pi$ times the square of the radius of the disk. The radius of each disk is the distance from the curve $y = \sqrt{\sin 4x}$ to the line $y = 1$, which is $1 - \sqrt{\sin 4x}$. The volume is thus $$V = \pi \int_{0}^{\frac{\pi}{4}} (1 - \sqrt{\sin 4x})^2 dx$$, because the region is bounded on the left by the $y$-axis ($x=0$) and on the right by the point where $\sin 4x = 1$, which occurs at $x = \frac{\pi}{4}$. Simplifying and solving this integral leads to the correct answer, which involves both $\pi$ and $\pi^2$ terms, indicating a subtraction of areas in the integral's solution.

Find the Volume of the Solid Generated by Revolving the Region