2 pts
Which choice below represents the area of the region that lies inside the second curve and outside the first curve where
\[
r_{1}=3 \sin \theta ; \quad r_{2}=\sqrt{3} \cos \theta
\]
$\frac{1}{2} \int_{\pi / 6}^{\pi / 2} r_{1}^{2} d \theta-\frac{1}{2} \int_{0}^{\pi / 6} r_{2}^{2} d \theta$
$\frac{1}{2} \int_{0}^{\pi / 6}\left(r_{2}^{2}-r_{1}^{2}\right) d \theta+\frac{1}{2} \int_{\pi / 2}^{\pi} r_{2}^{2} d \theta$
$\frac{1}{2} \int_{\pi / 6}^{\pi / 2} r_{2}^{2} d \theta-\frac{1}{2} \int_{0}^{\pi / 6} r_{1}^{2} d \theta$
$\frac{1}{2} \int_{\pi / 6}^{\pi} r_{1}^{2} d \theta-\frac{1}{2} \int_{\pi / 6}^{\pi / 2} r_{2}^{2} d \theta$

Step 1 ：First, we need to find the intersection points of the two curves. Set \(r_{1} = r_{2}\), we get \(3 \sin \theta = \sqrt{3} \cos \theta\). Solving this equation, we get \(\theta = \frac{\pi}{6}\).

Step 2 ：Next, we need to determine which curve is inside and which is outside in the region \(\theta \in [0, \frac{\pi}{6}]\). We can take a test point, for example, \(\theta = \frac{\pi}{12}\). Substituting \(\theta = \frac{\pi}{12}\) into \(r_{1}\) and \(r_{2}\), we get \(r_{1} = \frac{3\sqrt{2}}{2}\) and \(r_{2} = \sqrt{6}\). Since \(r_{1} > r_{2}\), the curve \(r_{1}\) is outside and the curve \(r_{2}\) is inside in the region \(\theta \in [0, \frac{\pi}{6}]\).

Step 3 ：Then, we need to determine which curve is inside and which is outside in the region \(\theta \in [\frac{\pi}{6}, \frac{\pi}{2}]\). We can take a test point, for example, \(\theta = \frac{\pi}{3}\). Substituting \(\theta = \frac{\pi}{3}\) into \(r_{1}\) and \(r_{2}\), we get \(r_{1} = \frac{3\sqrt{3}}{2}\) and \(r_{2} = 0\). Since \(r_{1} > r_{2}\), the curve \(r_{1}\) is outside and the curve \(r_{2}\) is inside in the region \(\theta \in [\frac{\pi}{6}, \frac{\pi}{2}]\).

Step 4 ：Finally, the area of the region that lies inside the second curve and outside the first curve is given by the integral \(\frac{1}{2} \int_{0}^{\pi / 6} r_{1}^{2} d \theta - \frac{1}{2} \int_{\pi / 6}^{\pi / 2} r_{2}^{2} d \theta\).