Solve the problem.
-Find the volume generated by revolving the curve $y = \cos 2 x$ about the $x$-axis, $0 \leq x \leq \pi / 24$
A) $\frac { \pi ^ { 2 } } { 48 } + \frac { \pi } { 24 }$
B) $\frac { \pi ^ { 2 } } { 48 } + \frac { \pi } { 16 }$
C) $\frac { \pi } { 48 } + \frac { 1 } { 16 }$
D) $\frac { \pi ^ { 2 } } { 48 }$

Question

Quizplus · Accepted Answer

We use the formula for the volume of a solid of revolution about the $x$-axis:

$$V = \int_{a}^{b} \pi [f(x)]^2 dx$$

Here, we have $f(x) = \cos 2x$ and we are revolving about the $x$-axis, so we need to find the limits of integration and square the function:

$$V = \int_{0}^{\pi / 24} \pi \cos^2 (2x) dx$$

We can use the identity $\cos^2\theta = \frac12 (1+\cos 2\theta)$ to simplify the integral:

\begin{align*}
V &= \int_{0}^{\pi / 24} \frac{\pi}{2} (1+\cos 4x) dx \
&= \frac{\pi}{2} \left[x+\frac{1}{4} \sin 4x \right]_{0}^{\pi / 24} \
&= \frac{\pi}{2} \left(\frac{\pi}{24} + \frac{1}{4} \sin \frac{\pi}{6} \right) \
&= \frac{\pi}{2} \left(\frac{\pi}{24} + \frac{1}{4} \cdot \frac12 \right) \
&= \frac{\pi^2}{48} + \frac{\pi}{16}
\end{align*}

Therefore, the answer is (B).

Solve the Problem y=cos⁡2xy = \cos 2 xy=cos2x About The xxx -Axis 0≤x≤π/240 \leq x \leq \pi / 240≤x≤π/24

Solve the Problem $y = \cos 2 x$ About The $x$ -Axis $0 \leq x \leq \pi / 24$