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) $\frac { 3 } { 4 } \pi ^ { 2 } - \frac { 1 } { 6 } \pi$
B) $\frac { 1 } { 4 } \pi ^ { 2 } - \frac { 1 } { 2 } \pi$
C) $\frac { 3 } { 4 } \pi - \frac { 1 } { 6 }$
D) $\frac { 1 } { 4 } \pi ^ { 2 } + \frac { 1 } { 2 } \pi$

Question

Quizplus · Accepted Answer

First, we need to sketch the region and the line of revolution. It will look like this:

[asy]
import graph;
import solids;
import intersection;
import math;

size(220);
currentprojection = perspective(6,3,2);

//Draw the region
path reg = graph(sin(2*x)^0.5,0,pi/2)--(pi/2,1)--(0,1)--cycle;
draw(surface(reg),white,nolight);

//Draw the line of revolution
real fL(real x) { return -1; }
real fR(real x) { return 0; }
draw(graph(fL,fR,0,pi/2),blue+1bp);

//Draw the axis labels
label("$x$",(pi/2,0,-0.5),SE);
label("$y$",(0,1.2,-0.5),E);
label("$-1$",(0.2,-1,0),N);
[/asy]

Note that the region is above the line of revolution, so it will be helpful to re-arrange the formula for the curve:
$$y = \sqrt{\sin 2x} \quad \Longrightarrow \quad \sin 2x = y^2 \quad \Longrightarrow \quad x = \frac{1}{2} \sin^{-1}(y^2)$$
Now, we can apply the formula for the volume of a solid of revolution using cylindrical shells. The radius of each shell is the distance from the axis of revolution to the corresponding point on the curve, which is $1+y$. The height of each shell is the arc length of the curve from $x$ to $x+dx$, which is $ds = \sqrt{1+(2\cos 2x)dx}$. Thus, the volume is given by:
\begin{align*}
V &= \int_0^{\pi/2} 2\pi(1+y)\sqrt{1+(2\cos 2x)dx} \
&= 2\pi \int_0^1 (1+y)\sqrt{1+4y^2}dy \quad (y = \sqrt{\sin 2x}) \
&= 2\pi \left( \frac{1}{2} \int_0^1 \sqrt{1+4y^2}dy + \int_0^1 y\sqrt{1+4y^2}dy ight) \
&= \pi \left( \frac{1}{2} \cdot \frac{1}{4} \left[(1+4y^2)^{3/2}ight]_0^1 + \frac{1}{3} \cdot \frac{1}{4} \left[(1+4y^2)^{3/2}ight]_0^1 ight) \
&= \frac{3}{4} \pi^{2} - \frac{1}{6} \pi
\end{align*}
Therefore, the answer is $\boxed{	extbf{(A)}}$.

Find the Volume of the Solid Generated by Revolving the Region