Answer the following:
(a) Setup, but do not evaluate, an integral to find the total length of the curve r = sin(5θ).
(b) Evaluate an integral to the find the area enclosed by one petal of the curve r = sin(5θ).

### (a) Setup the Integral for the Length of the Curve \( r = \sin(5\theta) \)

To find the total length of the curve given in polar coordinates, \( r = \sin(5\theta) \), we use the formula for the arc length in polar coordinates:

\[ L = \int_{a}^{b} \sqrt{ \left( \frac{dr}{d\theta} \right)^2 + r^2 } \, d\theta \]

First, we need to compute \(\frac{dr}{d\theta}\):

\[ r = \sin(5\theta) \]
\[ \frac{dr}{d\theta} = 5\cos(5\theta) \]

The total length of the curve over one full period is:

\[ L = \int_{0}^{2\pi} \sqrt{ \left( 5\cos(5\theta) \right)^2 + \left( \sin(5\theta) \right)^2 } \, d\theta \]

Simplify the integrand:

\[ L = \int_{0}^{2\pi} \sqrt{ 25\cos^2(5\theta) + \sin^2(5\theta) } \, d\theta \]

Therefore, the setup for the integral to find the total length of the curve is:

\[ L = \int_{0}^{2\pi} \sqrt{ 25\cos^2(5\theta) + \sin^2(5\theta) } \, d\theta \]

### (b) Evaluate the Integral to Find the Area Enclosed by One Petal of the Curve \( r = \sin(5\theta) \)

To find the area enclosed by one petal of the curve \( r = \sin(5\theta) \), we use the formula for the area in polar coordinates:

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

Each petal of the curve \( r = \sin(5\theta) \) occurs over the interval \( \theta \in \left[ 0, \frac{\pi}{5} \right] \). Therefore, for one petal, we have:

\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} (\sin(5\theta))^2 \, d\theta \]

To evaluate this integral, we use the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):

\[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} \left( \frac{1 - \cos(10\theta)}{2} \right) \, d\theta \]

\[ A = \frac{1}{4} \int_{0}^{\frac{\pi}{5}} (1 - \cos(10\theta)) \, d\theta \]

Now, we evaluate the integral:

\[ A = \frac{1}{4} \left[ \theta - \frac{\sin(10\theta)}{10} \right]_{0}^{\frac{\pi}{5}} \]

\[ A = \frac{1}{4} \left[ \frac{\pi}{5} - \frac{\sin(2\pi)}{10} - (0 - 0) \right] \]

Since \(\sin(2\pi) = 0\):

\[ A = \frac{1}{4} \left[ \frac{\pi}{5} \right] \]

\[ A = \frac{\pi}{20} \]

Therefore, the area enclosed by one petal of the curve \( r = \sin(5\theta) \) is:

\[ A = \frac{\pi}{20} \]

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Would you like further details or have any questions about these calculations? Here are five more questions you might be interested in:

1. How do you find the volume of a solid of revolution using integrals?
2. What is the method of partial fractions for integrating rational functions?
3. Can you explain the concept of a Taylor series expansion?
4. How do you evaluate improper integrals?
5. What are the applications of double integrals in physics?

**Tip:** When setting up integrals in polar coordinates, always consider the symmetry of the curve to simplify your calculations.

Explore integral calculus in polar coordinates with detailed solutions. Setup and evaluate integrals to find the total length and area enclosed by one petal of the curve r = sin(5θ).

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