曲线 $C$ 是一个半径为 2、圆心在原点的圆，其方程为 $x^2 + y^2 = 4$。直线 $L$ 的方程为 $\sqrt{3}y = x + 2$。下图显示了由 $C$ 和 $L$ 所围成的阴影区域 $A$。

1. Set up the area of region $A$ as the difference between the integral of the curve $C$ and the line $L$.

From the circle equation $x^2 + y^2 = 4$, the upper semicircle is $y = \sqrt{4-x^2}$.
From the line $L$, $y = \frac{x+2}{\sqrt{3}}$.
Area $A = \int_{-2}^{1} \sqrt{4-x^2}\ dx - \int_{-2}^{1} \frac{x+2}{\sqrt{3}}\ dx$

2. Apply the substitution $x = 2\sin\theta$ to the first integral and find the new limits.

Let $x = 2\sin\theta$. Then $\frac{dx}{d\theta} = 2\cos\theta \implies dx = 2\cos\theta\ d\theta$.
Find the new limits for $\theta$:
When $x = 1$, $2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$.
When $x = -2$, $2\sin\theta = -2 \implies \sin\theta = -1 \implies \theta = -\frac{\pi}{2}$.

3. Transform the first integral completely into terms of $\theta$.

$\int_{-2}^{1} \sqrt{4-x^2}\ dx = \int_{-\pi/2}^{\pi/6} \sqrt{4 - (2\sin\theta)^2} (2\cos\theta)\ d\theta$
$= \int_{-\pi/2}^{\pi/6} \sqrt{4(1-\sin^2\theta)} (2\cos\theta)\ d\theta$
$= \int_{-\pi/2}^{\pi/6} (2\cos\theta)(2\cos\theta)\ d\theta$
$= \int_{-\pi/2}^{\pi/6} 4\cos^2\theta\ d\theta$

4. Evaluate the second integral (the area under the line) to determine the exact constants $a, b,$ and $c$.

$\int_{-2}^{1} \frac{x+2}{\sqrt{3}}\ dx = \frac{1}{\sqrt{3}} \left[ \frac{x^2}{2} + 2x \right]_{-2}^{1}$
$= \frac{1}{\sqrt{3}} \left( \left(\frac{1}{2} + 2\right) - \left(\frac{4}{2} - 4\right) \right)$
$= \frac{1}{\sqrt{3}} \left( \frac{5}{2} - (-2) \right) = \frac{9}{2\sqrt{3}} = \frac{3\sqrt{3}}{2}$
Hence, Area $A = \int_{-\pi/2}^{\pi/6} 4\cos^2\theta\ d\theta - \frac{3\sqrt{3}}{2}$, where $a = -\frac{\pi}{2}$, $b = \frac{\pi}{6}$, and $c = \frac{3\sqrt{3}}{2}$.

5. Integrate $4\cos^2\theta$ using the double angle identity.

Using $\cos 2\theta = 2\cos^2\theta - 1$, we have $4\cos^2\theta = 2(1 + \cos 2\theta)$.
$\int_{-\pi/2}^{\pi/6} 2(1 + \cos 2\theta)\ d\theta = \left[ 2\theta + \sin 2\theta \right]_{-\pi/2}^{\pi/6}$

6. Evaluate the limits to find the exact area of region $A$.

$= \left( 2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{3}\right) \right) - \left( 2\left(-\frac{\pi}{2}\right) + \sin(-\pi) \right)$
$= \left( \frac{\pi}{3} + \frac{\sqrt{3}}{2} \right) - (-\pi + 0) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$
Exact Area $= \left(\frac{4\pi}{3} + \frac{\sqrt{3}}{2}\right) - \frac{3\sqrt{3}}{2} = \frac{4\pi}{3} - \sqrt{3}$

1. Express the equations of curve $C$ and line $L$ with $x$ as the subject.

For curve $C$: $x^2 + y^2 = 4 \implies x = \sqrt{4 - y^2}$ (taking the positive root as the region is in the first quadrant where $x > 0$).

For line $L$: $\sqrt{3}y = x + 2 \implies x = \sqrt{3}y - 2$.

2. Determine the $y$-coordinates of the intersections that bound Region $B$.

Region $B$ is bounded on the right by $x = 2$.

• Line $x = 2$ intersects curve $C$ ($x^2 + y^2 = 4$) at $2^2 + y^2 = 4 \implies y = 0$.
• Line $x = 2$ intersects line $L$ ($x = \sqrt{3}y - 2$) at $2 = \sqrt{3}y - 2 \implies \sqrt{3}y = 4 \implies y = \frac{4}{\sqrt{3}}$.
• As given from earlier parts, $C$ and $L$ intersect at $(1, \sqrt{3})$.

Thus, the region $B$ spans from $y = 0$ to $y = \frac{4}{\sqrt{3}}$.

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

When rotating about the $y$-axis, the volume is $V = \pi \int (x_{\text{outer}}^2 - x_{\text{inner}}^2) \, dy$.
The outer boundary is consistently $x = 2$.
The inner boundary changes at the intersection $y = \sqrt{3}$.

• For $0 \le y \le \sqrt{3}$, the inner boundary is the curve $C$.
• For $\sqrt{3} \le y \le \frac{4}{\sqrt{3}}$, the inner boundary is the line $L$.

Hence, the volume splits into two parts: $V = V_1 + V_2$.
$V = \pi \int_{0}^{\sqrt{3}} \left[ 2^2 - (\sqrt{4 - y^2})^2 \right] \, dy + \pi \int_{\sqrt{3}}^{4/\sqrt{3}} \left[ 2^2 - (\sqrt{3}y - 2)^2 \right] \, dy$

4. Evaluate the first volume integral, $V_1$.

$V_1 = \pi \int_{0}^{\sqrt{3}} \left( 4 - (4 - y^2) \right) \, dy$
$V_1 = \pi \int_{0}^{\sqrt{3}} y^2 \, dy$
$V_1 = \pi \left[ \frac{y^3}{3} \right]_{0}^{\sqrt{3}} = \pi \left( \frac{3\sqrt{3}}{3} \right) = \pi \sqrt{3}$

5. Evaluate the second volume integral, $V_2$.

$V_2 = \pi \int_{\sqrt{3}}^{4/\sqrt{3}} \left[ 4 - (3y^2 - 4\sqrt{3}y + 4) \right] \, dy$
$V_2 = \pi \int_{\sqrt{3}}^{4/\sqrt{3}} (4\sqrt{3}y - 3y^2) \, dy$
$V_2 = \pi \left[ 2\sqrt{3}y^2 - y^3 \right]_{\sqrt{3}}^{4/\sqrt{3}}$

Substitute the upper limit ($y = \frac{4}{\sqrt{3}}$):
$2\sqrt{3}\left(\frac{16}{3}\right) - \left(\frac{64}{3\sqrt{3}}\right) = \frac{32\sqrt{3}}{3} - \frac{64\sqrt{3}}{9} = \frac{96\sqrt{3} - 64\sqrt{3}}{9} = \frac{32\sqrt{3}}{9}$

Substitute the lower limit ($y = \sqrt{3}$):
$2\sqrt{3}(3) - (\sqrt{3})^3 = 6\sqrt{3} - 3\sqrt{3} = 3\sqrt{3} = \frac{27\sqrt{3}}{9}$

$V_2 = \pi \left( \frac{32\sqrt{3}}{9} - \frac{27\sqrt{3}}{9} \right) = \frac{5\sqrt{3}}{9}\pi$

6. Sum the volumes and compute the final numerical answer to two decimal places.

$V_{\text{total}} = V_1 + V_2 = \pi \sqrt{3} + \frac{5\sqrt{3}}{9}\pi = \frac{14\sqrt{3}}{9}\pi$
$V_{\text{total}} \approx 8.4636$
Rounding to 2 decimal places gives $8.46$.