曲线 $C$ 的方程为 $y = x\cos 2x$，其中 $0 \le x \le \pi$。

1. Set $y = 0$ to find the points of intersection with the $x$-axis.

$x \cos 2x = 0$

2. Solve for the first factor, $x = 0$.

$x = 0$

3. Solve for the second factor, $\cos 2x = 0$, within the given domain $0 \le x \le \pi$.

For $0 \le x \le \pi$, the range for $2x$ is $0 \le 2x \le 2\pi$.
$\cos 2x = 0$
$2x = \frac{\pi}{2}, \frac{3\pi}{2}$
$x = \frac{\pi}{4}, \frac{3\pi}{4}$

1. State the coordinates of the points where the curve crosses the axes.

From part (i), setting $y = 0$ for $0 \le x \le \pi$ yields $x = 0, \frac{\pi}{4}, \frac{3\pi}{4}$.
When $x = 0$, $y = 0$.
Thus, the curve meets or crosses the axes at $(0, 0)$, $\left(\frac{\pi}{4}, 0\right)$, and $\left(\frac{3\pi}{4}, 0\right)$.

2. Determine the behavior and endpoints to complete the sketch.

The endpoint is at $x = \pi$, where $y = \pi \cos(2\pi) = \pi$.
Using a graphing calculator to observe the shape over $0 \le x \le \pi$, the curve starts at $(0,0)$, increases to a local maximum, decreases to cross the $x$-axis at $\frac{\pi}{4}$, reaches a local minimum, increases to cross the $x$-axis at $\frac{3\pi}{4}$, and terminates at the point $(\pi, \pi)$.

1. Identify the interval where the integrand is negative and split the definite integral to resolve the modulus.

From part (i), $x\cos 2x = 0$ at $x = \frac{\pi}{4}$ and $x = \frac{3\pi}{4}$.
For $x \in \left[\frac{\pi}{4}, \frac{3\pi}{4}\right]$, $\cos 2x \le 0$, so $x\cos 2x \le 0$.
For $x \in \left[\frac{3\pi}{4}, \pi\right]$, $\cos 2x \ge 0$, so $x\cos 2x \ge 0$.
$\int_{\frac{\pi}{4}}^{\pi} |x\cos 2x|\,dx = \int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} -x\cos 2x \,dx + \int_{\frac{3\pi}{4}}^{\pi} x\cos 2x \,dx$

2. Perform integration by parts to find the indefinite integral of $x\cos 2x$.

Let $u = x \implies \frac{du}{dx} = 1$.
Let $\frac{dv}{dx} = \cos 2x \implies v = \frac{1}{2}\sin 2x$.
$\int x\cos 2x \,dx = x\left(\frac{1}{2}\sin 2x\right) - \int \frac{1}{2}\sin 2x \,dx$
$= \frac{1}{2}x\sin 2x - \left( -\frac{1}{4}\cos 2x \right) = \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x$

3. Evaluate the first definite integral over the interval $[rac{\pi}{4}, \frac{3\pi}{4}]$.

$\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} -x\cos 2x \,dx = -\left[ \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x \right]_{\frac{\pi}{4}}^{\frac{3\pi}{4}}$
$= -\left[ \left( \frac{1}{2}\left(\frac{3\pi}{4}\right)\sin\left(\frac{3\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{3\pi}{2}\right) \right) - \left( \frac{1}{2}\left(\frac{\pi}{4}\right)\sin\left(\frac{\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{\pi}{2}\right) \right) \right]$
$= -\left[ \left( \frac{3\pi}{8}(-1) + 0 \right) - \left( \frac{\pi}{8}(1) + 0 \right) \right]$
$= -\left( -\frac{3\pi}{8} - \frac{\pi}{8} \right) = \frac{4\pi}{8} = \frac{\pi}{2}$

4. Evaluate the second definite integral over the interval $[rac{3\pi}{4}, \pi]$ and sum the results.

$\int_{\frac{3\pi}{4}}^{\pi} x\cos 2x \,dx = \left[ \frac{1}{2}x\sin 2x + \frac{1}{4}\cos 2x \right]_{\frac{3\pi}{4}}^{\pi}$
$= \left( \frac{1}{2}(\pi)\sin(2\pi) + \frac{1}{4}\cos(2\pi) \right) - \left( \frac{1}{2}\left(\frac{3\pi}{4}\right)\sin\left(\frac{3\pi}{2}\right) + \frac{1}{4}\cos\left(\frac{3\pi}{2}\right) \right)$
$= \left( 0 + \frac{1}{4}(1) \right) - \left( -\frac{3\pi}{8} \right) = \frac{1}{4} + \frac{3\pi}{8}$
Total integral value:
$\frac{\pi}{2} + \left( \frac{1}{4} + \frac{3\pi}{8} \right) = \frac{4\pi}{8} + \frac{3\pi}{8} + \frac{1}{4} = \frac{7\pi}{8} + \frac{1}{4}$

1. Identify the correct region of integration and set up the volume integral.

From part (i), the $x$-intercepts of $C$ are $x = 0, \frac{\pi}{4}$, and $\frac{3\pi}{4}$.
The specific region bounded by the curve $C$, the $x$-axis, and the vertical line $x = \pi$ lies between the final intercept and $x = \pi$, which is the interval $\left[\frac{3\pi}{4}, \pi\right]$.
Setting up the volume of revolution about the $x$-axis:
$V = \pi \int_{\frac{3\pi}{4}}^{\pi} y^2 \,dx = \pi \int_{\frac{3\pi}{4}}^{\pi} (x\cos 2x)^2 \,dx$

2. Evaluate the definite integral using a graphing calculator.

Using a graphing calculator to evaluate the integral numerically:
$V \approx 10.465$
$V = 10.5 \text{ (to 3 s.f.)}$