The volume of the solid obtained by rotating the plane region enclosed by $y = \frac { \sin ( 2 x ) } { x } , x = \frac { \pi } { 2 } ,$ the y-axis and the x-axis about the y-axis is
A) $2 \pi$
B) $\pi$
C)2
D)1
E) $4 \pi$
F) $\frac { \pi } { 2 }$
G)4
H) $\frac { 1 } { 12 }$

Question

The volume of the solid obtained by rotating the plane region enclosed by $y = \frac { \sin ( 2 x ) } { x } , x = \frac { \pi } { 2 } ,$ the y-axis and the x-axis about the y-axis is
A) $2 \pi$ 
B) $\pi$ 
C)2
D)1
E) $4 \pi$ 
F) $\frac { \pi } { 2 }$ 
G)4
H) $\frac { 1 } { 12 }$

Quizplus · Accepted Answer

The volume of the solid obtained by rotating the given region around the y-axis can be calculated using the method of disks or cylindrical shells. In this case, using the method of cylindrical shells is more straightforward. The formula for the volume of a solid of revolution generated by rotating a function $f(x)$ about the y-axis from $x=a$ to $x=b$ is given by $V = 2\pi \int_{a}^{b} x f(x) dx$.Given $f(x) = \frac{\sin(2x)}{x}$, $a = 0$, and $b = \frac{\pi}{2}$, the volume $V$ is:$$V = 2\pi \int_{0}^{\frac{\pi}{2}} x \frac{\sin(2x)}{x} dx = 2\pi \int_{0}^{\frac{\pi}{2}} \sin(2x) dx$$Solving the integral:$$V = 2\pi [-\frac{1}{2}\cos(2x)]_{0}^{\frac{\pi}{2}} = 2\pi [-\frac{1}{2}\cos(\pi) + \frac{1}{2}\cos(0)] = 2\pi [-\frac{1}{2}(-1) + \frac{1}{2}(1)] = 2\pi$$Therefore, the volume of the solid is $2\pi$, which corresponds to choice A.

The Volume of the Solid Obtained by Rotating the Plane y=sin⁡(2x)x,x=π2,y = \frac { \sin ( 2 x ) } { x } , x = \frac { \pi } { 2 } ,y=xsin(2x)​,x=2π​,

The Volume of the Solid Obtained by Rotating the Plane $y = \frac { \sin ( 2 x ) } { x } , x = \frac { \pi } { 2 } ,$