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 $\sqrt{3}y = x + 2$。下方图示显示了由 C 和 L 所围成的阴影区域 A。

1. Set up the area integral as the difference between the area under the curve and the area under the line.

From the given intersection points $(-2,0)$ and $(1,\sqrt{3})$, the limits of integration are $x = -2$ and $x = 1$.
$\text{Area} = \int_{-2}^{1} \left( \sqrt{4-x^2} - \frac{x+2}{\sqrt{3}} \right) \, dx$
$= \int_{-2}^{1} \sqrt{4-x^2} \, dx - \int_{-2}^{1} \frac{x+2}{\sqrt{3}} \, dx$

2. Evaluate the integral for the line to find the exact constant $c$.

$c = \int_{-2}^{1} \frac{x+2}{\sqrt{3}} \, dx = \frac{1}{\sqrt{3}} \left[ \frac{(x+2)^2}{2} \right]_{-2}^{1}$
$= \frac{1}{\sqrt{3}} \left( \frac{3^2}{2} - 0 \right) = \frac{9}{2\sqrt{3}} = \frac{3\sqrt{3}}{2}$

3. Apply the substitution $x = 2\sin\theta$ to the integral for the curve to find $a$ and $b$.

$x = 2\sin\theta \implies dx = 2\cos\theta \, d\theta$
Change of limits:
When $x = -2$, $2\sin\theta = -2 \implies \sin\theta = -1 \implies \theta = -\frac{\pi}{2}$
When $x = 1$, $2\sin\theta = 1 \implies \sin\theta = \frac{1}{2} \implies \theta = \frac{\pi}{6}$
Substitute into the integral:
$\int_{-2}^{1} \sqrt{4-x^2} \, dx = \int_{-\pi/2}^{\pi/6} \sqrt{4 - 4\sin^2\theta} \, (2\cos\theta) \, d\theta$
$= \int_{-\pi/2}^{\pi/6} (2\cos\theta)(2\cos\theta) \, d\theta = \int_{-\pi/2}^{\pi/6} 4\cos^2\theta \, d\theta$
Comparing this with $\displaystyle\int_b^a 4\cos^2\theta \, d\theta - c$, we have exact constants $a = \frac{\pi}{6}$ and $b = -\frac{\pi}{2}$.

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

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

5. Evaluate the definite integral and subtract $c$ to find the exact area.

$= \left( 2\left(\frac{\pi}{6}\right) + \sin\left(\frac{\pi}{3}\right) \right) - \left( 2\left(-\frac{\pi}{2}\right) + \sin(-\pi) \right)$
$= \left( \frac{\pi}{3} + \frac{\sqrt{3}}{2} \right) - (-\pi + 0) = \frac{4\pi}{3} + \frac{\sqrt{3}}{2}$
Finally, subtract $c$ to find the exact area:
$\text{Area} = \int_{b}^{a} 4\cos^2\theta \, d\theta - c = \left( \frac{4\pi}{3} + \frac{\sqrt{3}}{2} \right) - \frac{3\sqrt{3}}{2}$
$= \frac{4\pi}{3} - \sqrt{3}$

1. Express the boundary equations in terms of $y$ and determine the limits of integration along the $y$-axis.

$x^2 + y^2 = 4 \implies x = \sqrt{4-y^2} \quad \text{(since $x \ge 0$ in Region B)}$
$\sqrt{3}y = x + 2 \implies x = \sqrt{3}y - 2$
The outer boundary is the vertical line $x = 2$. The inner boundary is defined by the curve $C$ from $y = 0$ to $y = \sqrt{3}$, and the line $L$ from $y = \sqrt{3}$ to $y = \frac{4}{\sqrt{3}}$.

2. Set up the volume integrals using the washer method for the two separate sections of Region B.

$V = V_1 + V_2$
$V_1 = \pi \int_{0}^{\sqrt{3}} \left[ 2^2 - (\sqrt{4-y^2})^2 \right] \,\mathrm{d}y$
$V_2 = \pi \int_{\sqrt{3}}^{4/\sqrt{3}} \left[ 2^2 - (\sqrt{3}y - 2)^2 \right] \,\mathrm{d}y$

3. Evaluate the definite integral for the first volume segment ($V_1$).

$V_1 = \pi \int_{0}^{\sqrt{3}} (4 - (4 - y^2)) \,\mathrm{d}y$
$V_1 = \pi \int_{0}^{\sqrt{3}} y^2 \,\mathrm{d}y = \pi \left[ \frac{y^3}{3} \right]_0^{\sqrt{3}}$
$V_1 = \pi \left( \frac{(\sqrt{3})^3}{3} - 0 \right) = \pi \sqrt{3}$

4. Evaluate the definite integral for the second volume segment ($V_2$).

$V_2 = \pi \int_{\sqrt{3}}^{4/\sqrt{3}} \left( 4 - (3y^2 - 4\sqrt{3}y + 4) \right) \,\mathrm{d}y$
$V_2 = \pi \int_{\sqrt{3}}^{4/\sqrt{3}} (4\sqrt{3}y - 3y^2) \,\mathrm{d}y$
$V_2 = \pi \left[ 2\sqrt{3}y^2 - y^3 \right]_{\sqrt{3}}^{4/\sqrt{3}}$
$V_2 = \pi \left( \left( 2\sqrt{3}\left(\frac{16}{3}\right) - \frac{64}{3\sqrt{3}} \right) - \left( 2\sqrt{3}(3) - 3\sqrt{3} \right) \right)$
$V_2 = \pi \left( \left( \frac{32\sqrt{3}}{3} - \frac{64\sqrt{3}}{9} \right) - 3\sqrt{3} \right)$
$V_2 = \pi \left( \frac{96\sqrt{3} - 64\sqrt{3}}{9} - \frac{27\sqrt{3}}{9} \right) = \frac{5\sqrt{3}\pi}{9}$

5. Sum the volumes and convert to a decimal value rounded to two decimal places.

$V = \pi\sqrt{3} + \frac{5\sqrt{3}\pi}{9} = \frac{14\sqrt{3}\pi}{9}$
$V \approx 8.46439 \dots \approx 8.46$