Find the area of the region that lies inside the first curve and outside the second curve. r = 3 cose, r = 1+ cose O a. A = 2π T O b. A=- OC. A = π O d. A = 3 TT 2 O e. A = 1/2
Find the area of the region that lies inside the first curve and outside the second curve. r = 3 cose, r = 1+ cose O a. A = 2π T O b. A=- OC. A = π O d. A = 3 TT 2 O e. A = 1/2
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Example Problem: Polar Coordinates and Area Between Curves
**Problem Statement:**
Find the area of the region that lies inside the first curve and outside the second curve.
**Given Curves:**
\[ r = 3\cos\theta \]
\[ r = 1 + \cos\theta \]
**Options:**
- (a) \( A = 2\pi \)
- (b) \( A = \frac{\pi}{4} \)
- (c) \( A = \pi \)
- (d) \( A = \frac{3\pi}{2} \)
- (e) \( A = \frac{\pi}{2} \)
This problem involves determining the area of a region defined by polar curves. Specifically, you must find the area that is enclosed within the curve \( r = 3\cos\theta \) and lies outside the curve \( r = 1 + \cos\theta \).
### Key Concepts
1. **Polar Coordinates:** A way to represent points in a plane using a radius and an angle.
2. **Area in Polar Coordinates:** For a polar curve \( r = f(\theta) \), the area from \( \theta = a \) to \( \theta = b \) is given by:
\[
A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 \, d\theta
\]
3. **Regions Between Curves:** To find the area between two curves \( r_1(\theta) \) and \( r_2(\theta) \), calculate the difference of their areas:
\[
A = \frac{1}{2} \int_{a}^{b} \left( [r_1(\theta)]^2 - [r_2(\theta)]^2 \right) d\theta
\]
### Approach:
1. **Find Intersection Points:** Determine the values of \( \theta \) where the two curves intersect by solving \( 3\cos\theta = 1 + \cos\theta \).
2. **Set Up Integral:** Identify the appropriate limits of integration and set up the integral for the area.
3. **Evaluate Integral:** Compute the integral to find the area.
Selecting the appropriate area formula and solving correctly will help you determine which of the given options is correct.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0d8d0c97-e66d-424a-9c52-d4c7d8464ad8%2F4154080f-0bf4-401a-b07b-a9b8e5be0b5c%2Fb1c3ryg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example Problem: Polar Coordinates and Area Between Curves
**Problem Statement:**
Find the area of the region that lies inside the first curve and outside the second curve.
**Given Curves:**
\[ r = 3\cos\theta \]
\[ r = 1 + \cos\theta \]
**Options:**
- (a) \( A = 2\pi \)
- (b) \( A = \frac{\pi}{4} \)
- (c) \( A = \pi \)
- (d) \( A = \frac{3\pi}{2} \)
- (e) \( A = \frac{\pi}{2} \)
This problem involves determining the area of a region defined by polar curves. Specifically, you must find the area that is enclosed within the curve \( r = 3\cos\theta \) and lies outside the curve \( r = 1 + \cos\theta \).
### Key Concepts
1. **Polar Coordinates:** A way to represent points in a plane using a radius and an angle.
2. **Area in Polar Coordinates:** For a polar curve \( r = f(\theta) \), the area from \( \theta = a \) to \( \theta = b \) is given by:
\[
A = \frac{1}{2} \int_{a}^{b} [f(\theta)]^2 \, d\theta
\]
3. **Regions Between Curves:** To find the area between two curves \( r_1(\theta) \) and \( r_2(\theta) \), calculate the difference of their areas:
\[
A = \frac{1}{2} \int_{a}^{b} \left( [r_1(\theta)]^2 - [r_2(\theta)]^2 \right) d\theta
\]
### Approach:
1. **Find Intersection Points:** Determine the values of \( \theta \) where the two curves intersect by solving \( 3\cos\theta = 1 + \cos\theta \).
2. **Set Up Integral:** Identify the appropriate limits of integration and set up the integral for the area.
3. **Evaluate Integral:** Compute the integral to find the area.
Selecting the appropriate area formula and solving correctly will help you determine which of the given options is correct.
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