The region under the curve $y = 3 \sin ^ { 2 } x , 0 \leq x \leq \pi$ is rotated about the $x$-axis. Find the volume of the resulting solid. Select the correct answer.
A) $\frac { 9 } { \pi ^ { 2 } }$
B) $\frac { 27 } { 8 } \pi ^ { 2 }$
C) $\frac { 3 } { \pi }$
D) $\frac { \pi ^ { 2 } } { 9 }$
E) $\frac { \pi } { 9 }$

Question

Quizplus · Accepted Answer

First, we need to find the equation of the curve obtained by rotating the given curve about the x-axis. The volume of the resulting solid can be found using the formula $\pi \int_{a}^{b} y^{2} dx$, where $a$ and $b$ are the limits of integration.

To find the equation of the curve, we need to consider a small element of the curve at $x$ with length $\Delta x$. The curve can be approximated as a rectangle with height $3\sin^{2}x$ and width $\Delta x$. When this rectangle is rotated about the x-axis, it generates a cylindrical shell with radius $y=3\sin^{2}x$ and height $\Delta x$. The volume of this shell is $2\pi y \Delta x$.

The total volume of the solid can be obtained by adding up the volumes of all such cylindrical shells between $0$ and $\pi$. This can be expressed as an integral:

$$\begin{aligned} V &= \int_{0}^{\pi} 2\pi y \Delta x \ &= 2\pi \int_{0}^{\pi} 3\sin^{2}x \Delta x \end{aligned}$$

To evaluate this integral, we can use the identity $\sin^{2}x = \frac{1}{2}(1-\cos2x)$. Then,

$$\begin{aligned} V &= 2\pi \int_{0}^{\pi} \frac{3}{2}(1-\cos2x) dx \ &= 3\pi \left[x - \frac{1}{2}\sin2x ight]_{0}^{\pi} \ &= 3\pi \left[\pi - 0 ight] \ &= 3\pi^{2} \end{aligned}$$

Hence, the correct answer is $\frac{27}{8}\pi^{2}$, which is option (B).

The Region Under the Curve y=3sin⁡2x,0≤x≤πy = 3 \sin ^ { 2 } x , 0 \leq x \leq \piy=3sin2x,0≤x≤π

The Region Under the Curve $y = 3 \sin ^ { 2 } x , 0 \leq x \leq \pi$