该图显示了方程为 $y = 2\sin^{-1}x$ 的曲线 $C$ 以及方程为 $y = \frac{8\pi}{3}x - \pi$ 的直线 $L$。$C$ 和 $L$ 相交于 $x = \frac{1}{2}$ 的点。 区域 $S$ 由 $y \le 2\sin^{-1}x$，$y \ge \frac{8\pi}{3}x - \pi$ 和 $x \ge 0$ 定义。 求将区域 $S$ 绕 $y$ 轴旋转 $2\pi$ 弧度所得几何体的精确体积。

1. Determine the $y$-coordinates of the intersection points and the intercepts.

When $x = \frac{1}{2}$,
$y = 2\sin^{-1}\left(\frac{1}{2}\right) = 2\left(\frac{\pi}{6}\right) = \frac{\pi}{3}$
For curve $C$, when $x = 0$, $y = 2\sin^{-1}(0) = 0$.
For line $L$, when $x = 0$, $y = -\pi$.

2. Express $x$ in terms of $y$ for both $C$ and $L$ to prepare for integration with respect to $y$.

For curve $C$:
$y = 2\sin^{-1}x \implies \sin^{-1}x = \frac{y}{2} \implies x = \sin\left(\frac{y}{2}\right)$
For line $L$:
$y = \frac{8\pi}{3}x - \pi \implies x = \frac{3}{8\pi}(y + \pi)$

3. Set up the integral for the volume generated by rotating the region about the $y$-axis.

The region is bounded by $L$ on the right, and by $C$ (for $y \ge 0$) and the $y$-axis (for $y < 0$) on the left.
The volume can be found by calculating the volume of the solid cone generated by $L$ and subtracting the solid volume generated by $C$:
$V = \pi \int_{-\pi}^{\frac{\pi}{3}} \left[ \frac{3}{8\pi}(y + \pi) \right]^2 \text{d}y - \pi \int_{0}^{\frac{\pi}{3}} \sin^2\left(\frac{y}{2}\right) \text{d}y$

4. Calculate the volume generated by rotating the line $L$ (which forms a cone).

$V_1 = \pi \int_{-\pi}^{\frac{\pi}{3}} \frac{9}{64\pi^2}(y + \pi)^2 \text{d}y$
$= \frac{9}{64\pi} \left[ \frac{(y + \pi)^3}{3} \right]_{-\pi}^{\frac{\pi}{3}}$
$= \frac{3}{64\pi} \left( \frac{4\pi}{3} \right)^3 = \frac{3}{64\pi} \left( \frac{64\pi^3}{27} \right) = \frac{\pi^2}{9}$
(Alternatively, using the cone formula: $V_1 = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi\left(\frac{1}{2}\right)^2\left(\frac{\pi}{3} - (-\pi)\right) = \frac{\pi^2}{9}$)

5. Calculate the volume generated by rotating the curve $C$, using the double angle formula.

$V_2 = \pi \int_{0}^{\frac{\pi}{3}} \sin^2\left(\frac{y}{2}\right) \text{d}y$
Using the identity $\sin^2\left(\frac{y}{2}\right) = \frac{1}{2}(1 - \cos y)$:
$V_2 = \pi \int_{0}^{\frac{\pi}{3}} \frac{1}{2}(1 - \cos y) \text{d}y$
$= \frac{\pi}{2} \left[ y - \sin y \right]_{0}^{\frac{\pi}{3}}$
$= \frac{\pi}{2} \left( \frac{\pi}{3} - \sin\frac{\pi}{3} \right)$
$= \frac{\pi}{2} \left( \frac{\pi}{3} - \frac{\sqrt{3}}{2} \right) = \frac{\pi^2}{6} - \frac{\sqrt{3}\pi}{4}$

6. Subtract the volumes to find the exact final volume.

$V = V_1 - V_2 = \frac{\pi^2}{9} - \left( \frac{\pi^2}{6} - \frac{\sqrt{3}\pi}{4} \right)$
$= \frac{2\pi^2}{18} - \frac{3\pi^2}{18} + \frac{\sqrt{3}\pi}{4}$
$= \frac{\sqrt{3}\pi}{4} - \frac{\pi^2}{18}$