A region bounded by f(x) = 2x + 2, y = x, and x = 2 is shown below. Find the volume of the solid formed by revolving the region about the x-axis. 8 7 6 5 V 4 3 2: 1 2 3 4 5 6 7 8 X
A region bounded by f(x) = 2x + 2, y = x, and x = 2 is shown below. Find the volume of the solid formed by revolving the region about the x-axis. 8 7 6 5 V 4 3 2: 1 2 3 4 5 6 7 8 X
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:
A region bounded by \( f(x) = 2x + 2 \), \( y = x \), and \( x = 2 \) is shown below. Find the volume of the solid formed by revolving the region about the x-axis.
### Graph Explanation:
The provided graph illustrates three lines on a Cartesian plane:
- **\( y = f(x) = 2x + 2 \)**:
This line has a positive slope, intersecting the y-axis at \( (0, 2) \). It is plotted with a green line extending upwards.
- **\( y = x \)**:
This line represents a linear function with a slope of 1, forming a 45-degree angle with both the x-axis and y-axis. It originates from the origin (0,0) and is also plotted with a green line extending upwards.
- **\( x = 2 \)**:
This is a vertical line crossing the x-axis at \( x = 2 \) and is plotted with a green line.
The shaded region in the graph lies between these three lines, forming a triangular area. The graph is bounded and the task involves calculating the volume formed when this region is revolved around the x-axis.
### Volume Calculation:
To find the volume of the solid formed by revolving the given region around the x-axis, we use the disk method. The volume \( V \) is given by:
\[ V = \pi \int_{a}^{b} [R(x)]^2 - [r(x)]^2 \, dx \]
Where:
- \( R(x) \) is the outer radius function
- \( r(x) \) is the inner radius function (if applicable, otherwise \( r(x)\) is 0).
In this case:
- \( R(x) = 2x + 2 \) (function tracing the upper boundary)
- \( r(x) = x \) (function tracing the lower boundary)
The bounds of integration \( a \) and \( b \) are from \( 0 \) to \( 2 \) (the x-coordinates of the region boundaries). Therefore:
\[ V = \pi \int_{0}^{2} [(2x + 2)^2 - (x)^2] \, dx \]
### Solving the Integral:
1. Expand the terms inside the integral:
\[ (2x + 2)^2 =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6bc5b610-4c42-40dc-b81a-6ff9e1fb2857%2F643f3758-1440-414f-9348-20ac3163d1a5%2F8k78q8r_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
A region bounded by \( f(x) = 2x + 2 \), \( y = x \), and \( x = 2 \) is shown below. Find the volume of the solid formed by revolving the region about the x-axis.
### Graph Explanation:
The provided graph illustrates three lines on a Cartesian plane:
- **\( y = f(x) = 2x + 2 \)**:
This line has a positive slope, intersecting the y-axis at \( (0, 2) \). It is plotted with a green line extending upwards.
- **\( y = x \)**:
This line represents a linear function with a slope of 1, forming a 45-degree angle with both the x-axis and y-axis. It originates from the origin (0,0) and is also plotted with a green line extending upwards.
- **\( x = 2 \)**:
This is a vertical line crossing the x-axis at \( x = 2 \) and is plotted with a green line.
The shaded region in the graph lies between these three lines, forming a triangular area. The graph is bounded and the task involves calculating the volume formed when this region is revolved around the x-axis.
### Volume Calculation:
To find the volume of the solid formed by revolving the given region around the x-axis, we use the disk method. The volume \( V \) is given by:
\[ V = \pi \int_{a}^{b} [R(x)]^2 - [r(x)]^2 \, dx \]
Where:
- \( R(x) \) is the outer radius function
- \( r(x) \) is the inner radius function (if applicable, otherwise \( r(x)\) is 0).
In this case:
- \( R(x) = 2x + 2 \) (function tracing the upper boundary)
- \( r(x) = x \) (function tracing the lower boundary)
The bounds of integration \( a \) and \( b \) are from \( 0 \) to \( 2 \) (the x-coordinates of the region boundaries). Therefore:
\[ V = \pi \int_{0}^{2} [(2x + 2)^2 - (x)^2] \, dx \]
### Solving the Integral:
1. Expand the terms inside the integral:
\[ (2x + 2)^2 =
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