Consider the two-dimensional surface (“lamina”) bounded by y = e^x , y = 0, and x = 1 and whose density is given by ρ(x, y) = y. a) Find the mass of this lamina. b) Find the coordinates (x, y) of its center of mass.
Consider the two-dimensional surface (“lamina”) bounded by y = e^x , y = 0, and x = 1 and whose density is given by ρ(x, y) = y. a) Find the mass of this lamina. b) Find the coordinates (x, y) of its center of mass.
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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Question
Consider the two-dimensional surface (“lamina”) bounded by y = e^x , y = 0, and x = 1 and whose density is given by ρ(x, y) = y.
a) Find the mass of this lamina.
b) Find the coordinates (x, y) of its center of mass.
![### Problem Statement
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**b) Find the coordinates \((\bar{x}, \bar{y})\) of its center of mass.**
---
This problem involves calculating the center of mass of an object or system. The center of mass is a critical concept in physics and engineering, representing the point where the mass of a system is considered to be concentrated. To find the center of mass coordinates, you will typically use the formulas:
\[ \bar{x} = \frac{\sum (x_i \cdot m_i)}{\sum m_i} \]
\[ \bar{y} = \frac{\sum (y_i \cdot m_i)}{\sum m_i} \]
where \(x_i\) and \(y_i\) are the coordinates of the individual masses \(m_i\) in the system. The total mass is denoted by \(\sum m_i\).
By solving these equations, you can determine the coordinates \((\bar{x}, \bar{y})\), which indicate the center of mass in the xy-plane.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85835193-c12a-47b1-84e7-c4f85c14c2c9%2F401a625c-47ce-440c-9445-8f56745a794c%2F6ll742d_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
---
**b) Find the coordinates \((\bar{x}, \bar{y})\) of its center of mass.**
---
This problem involves calculating the center of mass of an object or system. The center of mass is a critical concept in physics and engineering, representing the point where the mass of a system is considered to be concentrated. To find the center of mass coordinates, you will typically use the formulas:
\[ \bar{x} = \frac{\sum (x_i \cdot m_i)}{\sum m_i} \]
\[ \bar{y} = \frac{\sum (y_i \cdot m_i)}{\sum m_i} \]
where \(x_i\) and \(y_i\) are the coordinates of the individual masses \(m_i\) in the system. The total mass is denoted by \(\sum m_i\).
By solving these equations, you can determine the coordinates \((\bar{x}, \bar{y})\), which indicate the center of mass in the xy-plane.
![### Problem Statement
**2)** Consider the two-dimensional surface ("lamina") bounded by \( y = e^x \), \( y = 0 \), and \( x = 1 \) and whose density is given by \( \rho(x,y) = y \).
**a)** Find the mass of this lamina.
### Explanation
We need to find the mass of the lamina with a density function \( \rho(x,y) = y \). The lamina is bounded by the curve \( y = e^x \), the line \( y = 0 \), and the line \( x = 1 \) in the \( xy \)-plane.
To calculate the mass, we use the double integral of the density function over the region \( R \) bounded by the given curves and lines:
\[ \text{Mass} = \iint_R \rho(x,y) \, dA = \iint_R y \, dA \]
First, we determine the limits of integration:
- The vertical slices (\( y \)-values) will range from \( y = 0 \) to \( y = e^x \).
- The horizontal slices (\( x \)-values) will range from \( x = 0 \) to \( x = 1 \).
Thus, the double integral becomes:
\[ \text{Mass} = \int_{0}^{1} \int_{0}^{e^x} y \, dy \, dx \]
### Solution Steps
1. **Integrate with respect to \( y \):**
\[ \int_{0}^{e^x} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{e^x} = \frac{(e^x)^2}{2} - \frac{0^2}{2} = \frac{e^{2x}}{2} \]
2. **Integrate with respect to \( x \):**
\[ \int_{0}^{1} \frac{e^{2x}}{2} \, dx = \frac{1}{2} \int_{0}^{1} e^{2x} \, dx \]
To integrate \( e^{2x} \), use substitution:
\[ u = 2x \implies du = 2 \, dx \implies dx = \frac](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F85835193-c12a-47b1-84e7-c4f85c14c2c9%2F401a625c-47ce-440c-9445-8f56745a794c%2Fhdy3dlq_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
**2)** Consider the two-dimensional surface ("lamina") bounded by \( y = e^x \), \( y = 0 \), and \( x = 1 \) and whose density is given by \( \rho(x,y) = y \).
**a)** Find the mass of this lamina.
### Explanation
We need to find the mass of the lamina with a density function \( \rho(x,y) = y \). The lamina is bounded by the curve \( y = e^x \), the line \( y = 0 \), and the line \( x = 1 \) in the \( xy \)-plane.
To calculate the mass, we use the double integral of the density function over the region \( R \) bounded by the given curves and lines:
\[ \text{Mass} = \iint_R \rho(x,y) \, dA = \iint_R y \, dA \]
First, we determine the limits of integration:
- The vertical slices (\( y \)-values) will range from \( y = 0 \) to \( y = e^x \).
- The horizontal slices (\( x \)-values) will range from \( x = 0 \) to \( x = 1 \).
Thus, the double integral becomes:
\[ \text{Mass} = \int_{0}^{1} \int_{0}^{e^x} y \, dy \, dx \]
### Solution Steps
1. **Integrate with respect to \( y \):**
\[ \int_{0}^{e^x} y \, dy = \left[ \frac{y^2}{2} \right]_{0}^{e^x} = \frac{(e^x)^2}{2} - \frac{0^2}{2} = \frac{e^{2x}}{2} \]
2. **Integrate with respect to \( x \):**
\[ \int_{0}^{1} \frac{e^{2x}}{2} \, dx = \frac{1}{2} \int_{0}^{1} e^{2x} \, dx \]
To integrate \( e^{2x} \), use substitution:
\[ u = 2x \implies du = 2 \, dx \implies dx = \frac
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