To calculate the area of one loop of the polar curve $R = \sin(6\theta)$, we use the formula for the area enclosed by a polar curve:

$A = \frac{1}{2} \int_{\theta_1}^{\theta_2} R^2 \, d\theta$

Step 1: Determine the limits of integration

The curve $R = \sin(6\theta)$ completes one loop when $R$ returns to zero from zero. This occurs when: $\sin(6\theta) = 0$ $6\theta = n\pi, \quad n \in \mathbb{Z}$ Thus, the solutions are: $\theta = \frac{n\pi}{6}$ For one loop, we consider the interval from $\theta = 0$ to $\theta = \frac{\pi}{6}$.

Step 2: Set up the integral

The area of one loop is: $A = \frac{1}{2} \int_0^{\frac{\pi}{6}} \left(\sin(6\theta)\right)^2 d\theta$

Step 3: Simplify the integrand

Using the trigonometric identity: $\sin^2(x) = \frac{1 - \cos(2x)}{2},$ we have: $\sin^2(6\theta) = \frac{1 - \cos(12\theta)}{2}.$

Thus, the integral becomes: