Find the area of the region.
Two petals of $r=8 \sin (3 \theta)$

Step 1 ：The given equation is in polar coordinates. The area of a polar curve is given by the formula \(A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 d\theta\).

Step 2 ：The curve \(r=8 \sin (3 \theta)\) has 6 petals, but we only want the area of two petals. Since the petals are symmetric, we can find the area of one petal and multiply it by 2.

Step 3 ：To find the limits of integration, we need to find the values of \(\theta\) for which \(r=0\). Setting \(r=0\) gives \(\sin (3 \theta) = 0\), which has solutions \(\theta = 0, \frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}\).

Step 4 ：We can take \(\alpha = 0\) and \(\beta = \frac{\pi}{3}\) to get one petal, and then multiply the result by 2 to get the area of two petals.

Step 5 ：The area of one petal is approximately 5.33\(\pi\).

Step 6 ：So, the area of two petals is approximately 10.67\(\pi\).

Step 7 ：Final Answer: The area of the region is \(\boxed{10.67 \pi}\).