To find the area enclosed by one loop of the curve $r = 2 \sin(9\theta)$, we can use the formula for the area in polar coordinates:

$A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta$

Here, $r = 2 \sin(9\theta)$, so $r^2 = (2 \sin(9\theta))^2 = 4 \sin^2(9\theta)$. We need to find the appropriate bounds $\alpha$ and $\beta$ that define one loop of the curve.

Since the sine function completes one cycle when $9\theta$ goes from $0$ to $\pi$, this suggests that one loop of the curve corresponds to $\theta$ ranging from $0$ to $\frac{\pi}{9}$.

Now, the area is:

$A = \frac{1}{2} \int_0^{\frac{\pi}{9}} 4 \sin^2(9\theta) \, d\theta$

This simplifies to:

$A = 2 \int_0^{\frac{\pi}{9}} \sin^2(9\theta) \, d\theta$

We can use the trigonometric identity:

$\sin^2(x) = \frac{1 - \cos(2x)}{2}$

Substituting this into the integral:

$A = 2 \int_0^{\frac{\pi}{9}} \frac{1 - \cos(18\theta)}{2} \, d\theta$

This simplifies further to:

$A = \int_0^{\frac{\pi}{9}} (1 - \cos(18\theta)) \, d\theta$

Now, integrate:

$A = \left[ \theta - \frac{\sin(18\theta)}{18} \right]_0^{\frac{\pi}{9}}$

Evaluating this:

$A = \left( \frac{\pi}{9} - \frac{\sin(2\pi)}{18} \right) - \left( 0 - \frac{\sin(0)}{18} \right)$

Since $\sin(2\pi) = 0$ and $\sin(0) = 0$, we get:

$A = \frac{\pi}{9}$

Thus, the area enclosed by one loop of the curve is $\frac{\pi}{9}$.

Would you like more details on any step or have further questions?

Here are 5 related questions:

1. How do you determine the limits of integration in polar coordinates?
2. What is the geometric interpretation of polar curves like $r = 2\sin(9\theta)$?
3. How does the identity $\sin^2(x) = \frac{1 - \cos(2x)}{2}$ help simplify integrals involving trigonometric functions?
4. Can the area of more complex polar curves be computed using the same method?
5. What changes if the function involves $\cos(9\theta)$ instead of $\sin(9\theta)$?

Tip: For integrals of trigonometric functions, using identities to simplify the function can often make the integration much more straightforward.