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We are given the polar equation:

\[
r = 2 \sin(9\theta)
\]

and asked to find the area enclosed by one loop of the curve.

### Step 1: Formula for the area in polar coordinates

The formula for the area enclosed by a polar curve \( r = f(\theta) \) over a given interval is:

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

In this case, \( r = 2 \sin(9\theta) \), so we need to compute:

\[
A = \frac{1}{2} \int_{\alpha}^{\beta} [2 \sin(9\theta)]^2 \, d\theta
\]

### Step 2: Simplify the expression

First, simplify \( [2 \sin(9\theta)]^2 \):
\[
[2 \sin(9\theta)]^2 = 4 \sin^2(9\theta)
\]
Thus, the area formula becomes:
\[
A = 2 \int_{\alpha}^{\beta} \sin^2(9\theta) \, d\theta
\]

### Step 3: Determine the limits of integration

Since the curve is periodic, one loop of the curve corresponds to a full oscillation of the sine function. The sine function \( \sin(9\theta) \) completes one cycle (one loop) when \( 9\theta \) goes from \( 0 \) to \( \pi \), which corresponds to \( \theta \) going from \( 0 \) to \( \frac{\pi}{9} \).

Thus, the limits of integration are \( \alpha = 0 \) and \( \beta = \frac{\pi}{9} \).

### Step 4: Use a trigonometric identity

We use the identity:
\[
\sin^2(x) = \frac{1 - \cos(2x)}{2}
\]
to rewrite the integral:
\[
A = 2 \int_0^{\frac{\pi}{9}} \frac{1 - \cos(18\theta)}{2} \, d\theta
\]
Simplifying:
\[
A = \int_0^{\frac{\pi}{9}} (1 - \cos(18\theta)) \, d\theta
\]

### Step 5: Integrate

Now, integrate term by term:
\[
\int_0^{\frac{\pi}{9}} 1 \, d\theta = \left[ \theta \right]_0^{\frac{\pi}{9}} = \frac{\pi}{9}
\]
\[
\int_0^{\frac{\pi}{9}} \cos(18\theta) \, d\theta = \frac{1}{18} \sin(18\theta) \bigg|_0^{\frac{\pi}{9}} = \frac{1}{18} \sin(\pi) - \frac{1}{18} \sin(0) = 0
\]

Thus, the area simplifies to:
\[
A = \frac{\pi}{9}
\]

### Final Answer:

The area enclosed by one loop of the curve is:

\[
A = \frac{\pi}{9}
\]

Do you want more details or have any questions?

Here are five additional related questions to deepen your understanding:
1. How would the area formula change if the function involved \( \cos(9\theta) \) instead of \( \sin(9\theta) \)?
2. Can we calculate the area of multiple loops by adjusting the limits of integration?
3. How does the period of the sine function \( \sin(n\theta) \) affect the shape and number of loops in the curve?
4. What is the geometrical interpretation of a polar curve like \( r = 2 \sin(9\theta) \)?
5. How does the area formula in polar coordinates compare to finding areas in Cartesian coordinates?

**Tip**: When calculating areas in polar coordinates, always consider symmetries in the graph to simplify limits of integration, especially for periodic functions like sine or cosine.

Find the area of the region enclosed by one loop of the curve: r = 2 sin(9θ)

This problem involves calculating the area of the region enclosed by one loop of the polar curve r = 2 sin(9θ) using the area formula in polar coordinates. The solution includes steps to simplify the integral using trigonometric identities and setting appropriate limits of integration. Suitable for undergraduate calculus students.

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