 Find the area of the region enclosed by the rose r = 6cos(5).

To find the area of the region enclosed by the rose curve \( r = 6\cos(5\theta) \), we can use the formula for the area of a polar curve. A rose curve of the form \( r = a \cos(k\theta) \) or \( r = a \sin(k\theta) \) has \( 2k \) petals if \( k \) is odd and \( k \) petals if \( k \) is even. In our case, \( k = 5 \), which means the rose has 10 petals.

The formula for the area enclosed by one petal of the rose curve is:
\[
A = \frac{1}{2} \int_{\alpha}^{\beta} [r(\theta)]^2 \, d\theta
\]
Since the rose has 10 petals and each petal is identical, we can find the area of one petal and then multiply by 10 to get the total area.

### Steps to Find the Area

1. **Find the range for one petal**: Since \( k = 5 \), one petal is formed as \(\theta\) ranges from \( 0 \) to \( \frac{\pi}{5} \).

2. **Set up the integral for one petal**: The area of one petal is given by:
   \[
   A_{\text{petal}} = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} (6 \cos(5\theta))^2 \, d\theta
   \]

3. **Simplify the integral**:
   \[
   A_{\text{petal}} = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} 36 \cos^2(5\theta) \, d\theta
   \]
   \[
   = 18 \int_{0}^{\frac{\pi}{5}} \cos^2(5\theta) \, d\theta
   \]

4. **Use the double-angle identity for cosine**:
   \[
   \cos^2(5\theta) = \frac{1 + \cos(10\theta)}{2}
   \]
   So,
   \[
   A_{\text{petal}} = 18 \int_{0}^{\frac{\pi}{5}} \frac{1 + \cos(10\theta)}{2} \, d\theta
   \]
   \[
   = 9 \int_{0}^{\frac{\pi}{5}} (1 + \cos(10\theta)) \, d\theta
   \]

5. **Integrate**:
   \[
   9 \left[ \theta + \frac{\sin(10\theta)}{10} \right]_{0}^{\frac{\pi}{5}}
   \]
   Evaluate the integral at the bounds:
   \[
   = 9 \left( \left( \frac{\pi}{5} + \frac{\sin(2\pi)}{10} \right) - \left( 0 + \frac{\sin(0)}{10} \right) \right)
   \]
   \[
   = 9 \left( \frac{\pi}{5} + 0 - 0 \right)
   \]
   \[
   = 9 \cdot \frac{\pi}{5} = \frac{9\pi}{5}
   \]

6. **Calculate the total area**: Since there are 10 petals, multiply the area of one petal by 10:
   \[
   A_{\text{total}} = 10 \cdot \frac{9\pi}{5} = 18\pi
   \]

Therefore, the area of the region enclosed by the rose curve \( r = 6\cos(5\theta) \) is \( 18\pi \).

Would you like more details or have any questions?

Here are some related questions you might find interesting:

1. How does the number of petals change if the rose curve equation is \( r = a \cos(k\theta) \) with different values of \( k \)?
2. What happens to the shape of the rose curve if we change \( \cos \) to \( \sin \)?
3. Can you find the area enclosed by a rose curve with an even \( k \) value?
4. How do you derive the double-angle identity for trigonometric functions?
5. What are some other interesting properties of rose curves?
6. How do you graph rose curves by hand?
7. What is the significance of the parameter \( a \) in the rose curve equation?
8. How would you find the length of the curve of one petal of a rose curve?

**Tip**: When dealing with polar coordinates, always pay close attention to the periodicity of the trigonometric functions involved to properly set up your integrals.

Learn how to calculate the area enclosed by the polar rose curve r = 6cos(5θ) using integral calculus and trigonometric identities. Understand the concept of polar coordinates and how to find the area of a polar curve. Suitable for advanced high school students and calculus enthusiasts.

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