在同一坐标系中，草绘 $y = \sin 2x$ 和 $y = \cos x$ 在 $0 \le x \le \frac{\pi}{2}$ 上的图像，并写出曲线与坐标轴交点的精确坐标。 求不等式 $\sin 2x > \cos x$（其中 $0 \le x \le \frac{\pi}{2}$）的精确解。 由此，在不使用计算器的情况下，求出 $\int_{0}^{\frac{\pi}{2}} |\sin 2x - \cos x| \text{ d}x$ 的值。 由曲线 $y = \sin 2x$、$y = \cos x$ 以及 $y$ 轴（其中 $x \ge 0$）所围成的区域绕 $x$ 轴旋转 $2\pi$ 弧度。求所得旋转体的精确体积。

1. Sketch $y = \sin 2x$ for $0 \le x \le \frac{\pi}{2}$ and identify its intercepts.

The graph of $y = \sin 2x$ has a period of $\frac{2\pi}{2} = \pi$. For $0 \le x \le \frac{\pi}{2}$, it forms a single positive half-cycle, starting at $(0,0)$, reaching a maximum of $1$ at $x = \frac{\pi}{4}$, and returning to the $x$-axis at $(\frac{\pi}{2}, 0)$.
The intercepts are $(0,0)$ and $(\frac{\pi}{2},0)$.

2. Sketch $y = \cos x$ for $0 \le x \le \frac{\pi}{2}$ and identify its intercepts.

The graph of $y = \cos x$ has a period of $2\pi$. For $0 \le x \le \frac{\pi}{2}$, it forms a quarter-cycle, starting at a maximum on the $y$-axis at $(0,1)$ and strictly decreasing to cross the $x$-axis at $(\frac{\pi}{2}, 0)$.
The intercepts are $(0,1)$ and $(\frac{\pi}{2},0)$.

3. State the exact coordinates of the axial intercepts for both curves.

Combining the axis intersections of both sketched graphs, the exact coordinates where the curves cross the axes are $(0,0)$, $(0,1)$, and $(\frac{\pi}{2},0)$.

1. Find the exact intersection points of the curves $y = \sin 2x$ and $y = \cos x$ in the given domain.

To find where the curves intersect, set the equations equal and apply the double-angle identity:
$\sin 2x = \cos x$
$2 \sin x \cos x - \cos x = 0$
$\cos x (2 \sin x - 1) = 0$
This gives:
$\cos x = 0 \quad \text{or} \quad \sin x = \frac{1}{2}$
For $0 \le x \le \frac{\pi}{2}$, the solutions are:
$x = \frac{\pi}{2} \quad \text{and} \quad x = \frac{\pi}{6}$

2. Determine the interval satisfying the strict inequality $\sin 2x > \cos x$.

The inequality requires the curve $y = \sin 2x$ to be strictly above $y = \cos x$.
Testing a value between $x = \frac{\pi}{6}$ and $x = \frac{\pi}{2}$, such as $x = \frac{\pi}{4}$:
$\sin\left(2 \cdot \frac{\pi}{4}\right) = \sin \frac{\pi}{2} = 1$
$\cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \approx 0.707$
Since $1 > \frac{\sqrt{2}}{2}$, the inequality holds in this interval. Because it is a strict inequality, the endpoints where the curves intersect are not included.
$\frac{\pi}{6} < x < \frac{\pi}{2}$

1. Determine the split point for the absolute value function using the results from part (ii).

From part (ii), $\sin 2x = \cos x \implies \cos x(2\sin x - 1) = 0 \implies x = \frac{\pi}{6}$ or $x = \frac{\pi}{2}$.
We know $\sin 2x \le \cos x$ for $0 \le x \le \frac{\pi}{6}$, and $\sin 2x > \cos x$ for $\frac{\pi}{6} < x < \frac{\pi}{2}$.
Therefore, we split the absolute value integral:
$\int_{0}^{\frac{\pi}{2}} |\sin 2x - \cos x| \text{ d}x = \int_{0}^{\frac{\pi}{6}} (\cos x - \sin 2x) \text{ d}x + \int_{\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin 2x - \cos x) \text{ d}x$

2. Integrate both expressions with respect to $x$.

$= \left[ \sin x + \frac{1}{2}\cos 2x \right]_{0}^{\frac{\pi}{6}} + \left[ -\frac{1}{2}\cos 2x - \sin x \right]_{\frac{\pi}{6}}^{\frac{\pi}{2}}$

3. Substitute the exact limits and evaluate without using a calculator.

For the first integral:
$\left(\sin \frac{\pi}{6} + \frac{1}{2}\cos \frac{\pi}{3}\right) - \left(\sin 0 + \frac{1}{2}\cos 0\right) = \left(\frac{1}{2} + \frac{1}{2}\left(\frac{1}{2}\right)\right) - \left(0 + \frac{1}{2}\right) = \frac{3}{4} - \frac{1}{2} = \frac{1}{4}$

For the second integral:
$\left(-\frac{1}{2}\cos \pi - \sin \frac{\pi}{2}\right) - \left(-\frac{1}{2}\cos \frac{\pi}{3} - \sin \frac{\pi}{6}\right) = \left(\frac{1}{2} - 1\right) - \left(-\frac{1}{4} - \frac{1}{2}\right) = -\frac{1}{2} + \frac{3}{4} = \frac{1}{4}$

Total sum:
$= \frac{1}{4} + \frac{1}{4} = \frac{1}{2}$

1. Determine the limits of integration by finding the point of intersection between the two curves for $x \ge 0$.

$\sin 2x = \cos x$
$2 \sin x \cos x - \cos x = 0$
$\cos x (2 \sin x - 1) = 0$
Since we are considering the region bounded by the $y$-axis ($x = 0$), we look for the first intersection point.
$\sin x = \frac{1}{2} \implies x = \frac{\pi}{6}$
Thus, the region lies between $x = 0$ and $x = \frac{\pi}{6}$.

2. Set up the definite integral for the volume of revolution using the washer method.

For $0 \le x \le \frac{\pi}{6}$, the curve $y = \cos x$ lies above $y = \sin 2x$.
The volume $V$ generated is the difference between the volumes of revolution of the upper and lower curves:
$V = \pi \int_{0}^{\frac{\pi}{6}} \left( (\cos x)^2 - (\sin 2x)^2 \right) \text{ d}x$

3. Use double-angle trigonometric identities to linearise the integrand.

Applying the identities $\cos^2 A = \frac{1 + \cos 2A}{2}$ and $\sin^2 A = \frac{1 - \cos 2A}{2}$:
$\cos^2 x - \sin^2 2x = \frac{1 + \cos 2x}{2} - \frac{1 - \cos 4x}{2}$
$= \frac{1}{2}(1 + \cos 2x - 1 + \cos 4x)$
$= \frac{1}{2}(\cos 2x + \cos 4x)$
Substituting this back into the integral:
$V = \frac{\pi}{2} \int_{0}^{\frac{\pi}{6}} (\cos 2x + \cos 4x) \text{ d}x$

4. Evaluate the definite integral to find the exact volume.

$V = \frac{\pi}{2} \left[ \frac{1}{2}\sin 2x + \frac{1}{4}\sin 4x \right]_{0}^{\frac{\pi}{6}}$
$V = \frac{\pi}{2} \left( \left( \frac{1}{2}\sin\frac{\pi}{3} + \frac{1}{4}\sin\frac{2\pi}{3} \right) - (0 + 0) \right)$
$V = \frac{\pi}{2} \left( \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) + \frac{1}{4}\left(\frac{\sqrt{3}}{2}\right) \right)$
$V = \frac{\pi}{2} \left( \frac{\sqrt{3}}{4} + \frac{\sqrt{3}}{8} \right)$
$V = \frac{\pi}{2} \left( \frac{3\sqrt{3}}{8} \right) = \frac{3\sqrt{3}\pi}{16}$