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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.

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

Discover how to calculate the area enclosed by one loop of the polar curve r = 2sin(9θ) using polar integration. Learn about key mathematical concepts like polar coordinates, integration of trigonometric functions, and trigonometric identities, suitable for advanced high school or undergraduate math students.

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