Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. y = 6-6x², y = 0 V= Sketch the region. y -1.5 -10 -0.5 0.5 1.0 1.5 -0.5 0.5 1.0 1.5 -2 Sof X -1.5 -1.0 -0.5 0.5 1.0 1.5 -0.5 41 2 -2 -6- 0.5 .0 1.5
Find the volume of the solid obtained by rotating the region bounded by the given curves about the x-axis. y = 6-6x², y = 0 V= Sketch the region. y -1.5 -10 -0.5 0.5 1.0 1.5 -0.5 0.5 1.0 1.5 -2 Sof X -1.5 -1.0 -0.5 0.5 1.0 1.5 -0.5 41 2 -2 -6- 0.5 .0 1.5
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...
Related questions
Question
100%
![### Volume of Solid of Revolution
To find the volume of the solid obtained by rotating the region bounded by the given curves about the \( x \)-axis, follow these steps:
Consider the region bounded by:
\[ y = 6 - 6x^2, \qquad y = 0 \]
The volume \( V \) of the solid of revolution can be computed using the formula:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
where \( f(x) = 6 - 6x^2 \).
#### Step-by-Step Process
1. **Identify the Intersection Points**:
To find the limits of integration \( a \) and \( b \), set \( 6 - 6x^2 = 0 \):
\[ 6 - 6x^2 = 0 \]
\[ 6 = 6x^2 \]
\[ x^2 = 1 \]
\[ x = \pm 1 \]
Therefore, the bounds of integration are from \( x = -1 \) to \( x = 1 \).
2. **Set Up the Integral**:
The function \( f(x) \) which we are rotating about the x-axis is \( 6 - 6x^2 \).
The volume \( V \) is given by:
\[ V = \pi \int_{-1}^{1} (6 - 6x^2)^2 \, dx \]
3. **Solve the Integral**:
The integral can be split and then solved as follows:
\[ V = \pi \int_{-1}^{1} (6 - 6x^2)^2 \, dx \]
#### Graphical Representation:
The images below represent different stages of visual understanding of the problem:
1. **Original Region**:
- The leftmost graph shows the original region bounded by \( y = 6 - 6x^2 \) and \( y = 0 \). This is the region that will be rotated around the x-axis.
2. **Region After Rotation**:
- The second graph from the left demonstrates the corresponding region after it has been rotated by 180 degrees around the x-axis, forming a solid of revolution.
3. **Symmetric View**:
- The third graph depicts the rotated region emphasizing the symmetry of the](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd001228d-db5b-4f7d-bdcc-5b8b3535f34a%2F41fc9848-44ef-4877-9683-5ac84f2ac552%2Fih6dnn7_processed.png&w=3840&q=75)
Transcribed Image Text:### Volume of Solid of Revolution
To find the volume of the solid obtained by rotating the region bounded by the given curves about the \( x \)-axis, follow these steps:
Consider the region bounded by:
\[ y = 6 - 6x^2, \qquad y = 0 \]
The volume \( V \) of the solid of revolution can be computed using the formula:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
where \( f(x) = 6 - 6x^2 \).
#### Step-by-Step Process
1. **Identify the Intersection Points**:
To find the limits of integration \( a \) and \( b \), set \( 6 - 6x^2 = 0 \):
\[ 6 - 6x^2 = 0 \]
\[ 6 = 6x^2 \]
\[ x^2 = 1 \]
\[ x = \pm 1 \]
Therefore, the bounds of integration are from \( x = -1 \) to \( x = 1 \).
2. **Set Up the Integral**:
The function \( f(x) \) which we are rotating about the x-axis is \( 6 - 6x^2 \).
The volume \( V \) is given by:
\[ V = \pi \int_{-1}^{1} (6 - 6x^2)^2 \, dx \]
3. **Solve the Integral**:
The integral can be split and then solved as follows:
\[ V = \pi \int_{-1}^{1} (6 - 6x^2)^2 \, dx \]
#### Graphical Representation:
The images below represent different stages of visual understanding of the problem:
1. **Original Region**:
- The leftmost graph shows the original region bounded by \( y = 6 - 6x^2 \) and \( y = 0 \). This is the region that will be rotated around the x-axis.
2. **Region After Rotation**:
- The second graph from the left demonstrates the corresponding region after it has been rotated by 180 degrees around the x-axis, forming a solid of revolution.
3. **Symmetric View**:
- The third graph depicts the rotated region emphasizing the symmetry of the
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
This is a popular solution!
Trending now
This is a popular solution!
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781285741550
Author:
James Stewart
Publisher:
Cengage Learning
Thomas' Calculus (14th Edition)
Calculus
ISBN:
9780134438986
Author:
Joel R. Hass, Christopher E. Heil, Maurice D. Weir
Publisher:
PEARSON
Calculus: Early Transcendentals (3rd Edition)
Calculus
ISBN:
9780134763644
Author:
William L. Briggs, Lyle Cochran, Bernard Gillett, Eric Schulz
Publisher:
PEARSON
Calculus: Early Transcendentals
Calculus
ISBN:
9781319050740
Author:
Jon Rogawski, Colin Adams, Robert Franzosa
Publisher:
W. H. Freeman
Calculus: Early Transcendental Functions
Calculus
ISBN:
9781337552516
Author:
Ron Larson, Bruce H. Edwards
Publisher:
Cengage Learning