Find the area enclosed by one leaf of the rose r= 9 cos (50).
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
"Find the area enclosed by one leaf of the rose \( r = 9 \cos (5\theta) \)."
**Solution Overview:**
To determine the area enclosed by one leaf of a rose curve given in polar coordinates, \( r = 9 \cos (5\theta) \), you use the formula for the area in polar coordinates:
\[
A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta
\]
For a rose curve of the form \( r = a \cos(n\theta) \), each leaf occurs from \( \theta = 0 \) to \( \theta = \frac{\pi}{n} \).
In this specific case, \( r = 9 \cos(5\theta) \), the parameter \( a \) is 9 and \( n \) is 5. Thus, the integral bounds for one leaf will range from \( \alpha = 0 \) to \( \beta = \frac{\pi}{5} \).
**Detailed Steps:**
1. **Square the function:**
\[
r^2 = (9 \cos(5\theta))^2 = 81 \cos^2(5\theta)
\]
2. **Set up the integral for the area:**
\[
A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} 81 \cos^2(5\theta) \, d\theta
\]
3. **Simplify the integral:**
Use the trigonometric identity:
\[
\cos^2(x) = \frac{1 + \cos(2x)}{2}
\]
So,
\[
81 \cos^2(5\theta) = 81 \cdot \frac{1 + \cos(10\theta)}{2} = \frac{81}{2} (1 + \cos(10\theta))
\]
4. **Integrate:**
\[
A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} \frac{81}{2} (1 + \cos(10\theta)) \, d\theta
\]
\[
A = \frac{81}{4} \int_{0}^{\frac{\pi}{5}} (1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb21cc1b3-0a4b-474e-87d5-2a9f16683342%2Fcf2eaf88-b867-46b9-815f-406bf38405b2%2F4dajlgs_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
"Find the area enclosed by one leaf of the rose \( r = 9 \cos (5\theta) \)."
**Solution Overview:**
To determine the area enclosed by one leaf of a rose curve given in polar coordinates, \( r = 9 \cos (5\theta) \), you use the formula for the area in polar coordinates:
\[
A = \frac{1}{2} \int_{\alpha}^{\beta} r^2 \, d\theta
\]
For a rose curve of the form \( r = a \cos(n\theta) \), each leaf occurs from \( \theta = 0 \) to \( \theta = \frac{\pi}{n} \).
In this specific case, \( r = 9 \cos(5\theta) \), the parameter \( a \) is 9 and \( n \) is 5. Thus, the integral bounds for one leaf will range from \( \alpha = 0 \) to \( \beta = \frac{\pi}{5} \).
**Detailed Steps:**
1. **Square the function:**
\[
r^2 = (9 \cos(5\theta))^2 = 81 \cos^2(5\theta)
\]
2. **Set up the integral for the area:**
\[
A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} 81 \cos^2(5\theta) \, d\theta
\]
3. **Simplify the integral:**
Use the trigonometric identity:
\[
\cos^2(x) = \frac{1 + \cos(2x)}{2}
\]
So,
\[
81 \cos^2(5\theta) = 81 \cdot \frac{1 + \cos(10\theta)}{2} = \frac{81}{2} (1 + \cos(10\theta))
\]
4. **Integrate:**
\[
A = \frac{1}{2} \int_{0}^{\frac{\pi}{5}} \frac{81}{2} (1 + \cos(10\theta)) \, d\theta
\]
\[
A = \frac{81}{4} \int_{0}^{\frac{\pi}{5}} (1
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