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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![### Problem Statement
Find the mass of a lamina bounded by \( y = 5\sqrt{x} \), \( x = 1 \), and \( y = 0 \) with the density function \( \sigma(x,y) = x + 2 \).
### Explanation
To solve this problem, you need to integrate the density function over the region bounded by the given curves:
1. **Boundary Curves**:
- **Curve 1**: \(y = 5\sqrt{x} \)
- **Curve 2**: \( x = 1 \)
- **Curve 3**: \( y = 0 \) (the x-axis)
2. **Density Function**:
\[ \sigma(x,y) = x + 2 \]
### Solution
To find the mass of the lamina, we will integrate the density function over the bounded region.
First, express the boundaries in terms of \( x \):
- For \( y = 5\sqrt{x} \), solving for \( x \) gives \( x = \left(\frac{y}{5}\right)^2 \).
The region is bounded horizontally by \( x \) from \( 0 \) to \( 1 \), and vertically by \( y \) from \( 0 \) to \( 5\sqrt{1} = 5 \).
Thus, in rectangular form, the boundaries translated to limits of integration are:
- \( x \) ranges from 0 to 1.
- \( y \) ranges from 0 to \( 5\sqrt{x} \).
The mass \( M \) of the lamina can be found using the double integral:
\[ M = \int_{0}^{1} \int_{0}^{5\sqrt{x}} (x + 2) \, dy \, dx \]
1. **Integrate with respect to \( y \)**:
\[ \int_{0}^{5\sqrt{x}} (x + 2) \, dy = (x + 2) \int_{0}^{5\sqrt{x}} dy = (x + 2) \cdot 5\sqrt{x} = 5(x + 2)\sqrt{x} \]
2. **Integrate the result with respect to \( x \)**:
\[ M = \int_{0}^{1} 5(x + 2)\sqrt{x} \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F74b6836e-7075-4604-8b7e-b1441272473c%2F286b8b7c-bbd9-431e-a873-7c5d0b458aa0%2F4lnmtzg_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement
Find the mass of a lamina bounded by \( y = 5\sqrt{x} \), \( x = 1 \), and \( y = 0 \) with the density function \( \sigma(x,y) = x + 2 \).
### Explanation
To solve this problem, you need to integrate the density function over the region bounded by the given curves:
1. **Boundary Curves**:
- **Curve 1**: \(y = 5\sqrt{x} \)
- **Curve 2**: \( x = 1 \)
- **Curve 3**: \( y = 0 \) (the x-axis)
2. **Density Function**:
\[ \sigma(x,y) = x + 2 \]
### Solution
To find the mass of the lamina, we will integrate the density function over the bounded region.
First, express the boundaries in terms of \( x \):
- For \( y = 5\sqrt{x} \), solving for \( x \) gives \( x = \left(\frac{y}{5}\right)^2 \).
The region is bounded horizontally by \( x \) from \( 0 \) to \( 1 \), and vertically by \( y \) from \( 0 \) to \( 5\sqrt{1} = 5 \).
Thus, in rectangular form, the boundaries translated to limits of integration are:
- \( x \) ranges from 0 to 1.
- \( y \) ranges from 0 to \( 5\sqrt{x} \).
The mass \( M \) of the lamina can be found using the double integral:
\[ M = \int_{0}^{1} \int_{0}^{5\sqrt{x}} (x + 2) \, dy \, dx \]
1. **Integrate with respect to \( y \)**:
\[ \int_{0}^{5\sqrt{x}} (x + 2) \, dy = (x + 2) \int_{0}^{5\sqrt{x}} dy = (x + 2) \cdot 5\sqrt{x} = 5(x + 2)\sqrt{x} \]
2. **Integrate the result with respect to \( x \)**:
\[ M = \int_{0}^{1} 5(x + 2)\sqrt{x} \
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