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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I am struggling solving this question. The answer is (1/2, 0) and am not sure how they got 0 as the y-coordinate. Equations would definitely be helpful in the answer!
![**Problem 50: Center of Mass of a Region Bounded by Curves**
**Objective:**
Determine the center of mass of the region bounded by the curves \( y = x + 2 \) and \( y = x^2 - 4 \).
**Step-by-Step Approach:**
1. **Plotting the Curves:**
- The first curve, \( y = x + 2 \), is a straight line. It has a slope of 1 and a y-intercept at (0, 2).
- The second curve, \( y = x^2 - 4 \), is a parabola opening upwards, with its vertex at (0, -4).
2. **Finding Points of Intersection:**
- To find the points where the curves intersect, solve \( x + 2 = x^2 - 4 \):
\[
x^2 - x - 6 = 0
\]
This factors to:
\[
(x - 3)(x + 2) = 0
\]
Therefore, the points of intersection are:
\[
x = 3 \quad \text{and} \quad x = -2
\]
3. **Region Bounded by Curves:**
- The region of interest is between \( x = -2 \) and \( x = 3 \).
- The line \( y = x + 2 \) is above the parabola \( y = x^2 - 4 \) in this interval.
4. **Formulas for Center of Mass:**
The coordinates of the center of mass \((\bar{x}, \bar{y})\) for the region \(R\) are:
\[
\bar{x} = \frac{1}{A} \int_a^b x \left[ f(x) - g(x) \right] \, dx
\]
\[
\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} \left( f(x)^2 - g(x)^2 \right) \, dx
\]
where \( A \) is the area of the region \( R \):
\[
A = \int_a^b \left[ f(x) - g(x) \right](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc061ccfc-c5c1-460b-8498-b5bc9ae72844%2Ffcdf15ab-4422-4d53-b7d4-e25156b7ab5c%2Fgg7vr2_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 50: Center of Mass of a Region Bounded by Curves**
**Objective:**
Determine the center of mass of the region bounded by the curves \( y = x + 2 \) and \( y = x^2 - 4 \).
**Step-by-Step Approach:**
1. **Plotting the Curves:**
- The first curve, \( y = x + 2 \), is a straight line. It has a slope of 1 and a y-intercept at (0, 2).
- The second curve, \( y = x^2 - 4 \), is a parabola opening upwards, with its vertex at (0, -4).
2. **Finding Points of Intersection:**
- To find the points where the curves intersect, solve \( x + 2 = x^2 - 4 \):
\[
x^2 - x - 6 = 0
\]
This factors to:
\[
(x - 3)(x + 2) = 0
\]
Therefore, the points of intersection are:
\[
x = 3 \quad \text{and} \quad x = -2
\]
3. **Region Bounded by Curves:**
- The region of interest is between \( x = -2 \) and \( x = 3 \).
- The line \( y = x + 2 \) is above the parabola \( y = x^2 - 4 \) in this interval.
4. **Formulas for Center of Mass:**
The coordinates of the center of mass \((\bar{x}, \bar{y})\) for the region \(R\) are:
\[
\bar{x} = \frac{1}{A} \int_a^b x \left[ f(x) - g(x) \right] \, dx
\]
\[
\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} \left( f(x)^2 - g(x)^2 \right) \, dx
\]
where \( A \) is the area of the region \( R \):
\[
A = \int_a^b \left[ f(x) - g(x) \right
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