Find the volume of the solid generated by rotating the region enclosed by y = x + 3, y = 0, x = 1, and x = 4 about the x-axis. Do not simplify your answer. y = x + 3 I
Find the volume of the solid generated by rotating the region enclosed by y = x + 3, y = 0, x = 1, and x = 4 about the x-axis. Do not simplify your answer. y = x + 3 I
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter8: Further Techniques And Applications Of Integration
Section8.3: Volume And Average Value
Problem 22E
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Question
![**Topic: Calculating the Volume of Solid of Revolution**
**Problem Statement:**
Find the volume of the solid generated by rotating the region enclosed by \( y = x + 3 \), \( y = 0 \), \( x = 1 \), and \( x = 4 \) about the x-axis. Do not simplify your answer.
**Graphical Representation:**
The provided diagram illustrates the region enclosed by the following boundaries:
- The line \( y = x + 3 \), which is depicted in blue.
- The horizontal line \( y = 0 \) (the x-axis), shown in black.
- Two vertical lines representing \( x = 1 \) and \( x = 4 \), marked in red.
The shaded gray region indicates the area that will be revolved around the x-axis to form the solid.
**Explanation and Steps to Find the Volume:**
To find the volume of the solid formed by rotating the enclosed region around the x-axis, we use the disk method. The volume \( V \) is given by the integral:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
Here, the function \( f(x) = x + 3 \), and the limits of integration are from \( x = 1 \) to \( x = 4 \).
1. **Identify the function and limits:**
- \( f(x) = x + 3 \)
- Lower limit \( a = 1 \)
- Upper limit \( b = 4 \)
2. **Set up the integral:**
\[ V = \pi \int_{1}^{4} (x + 3)^2 \, dx \]
3. **Perform the integration:**
- Expand the integrand first:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
- Integrate each term separately:
\[ V = \pi \int_{1}^{4} (x^2 + 6x + 9) \, dx \]
\[ V = \pi \left[ \int_{1}^{4} x^2 \, dx + \int_{1}^{4} 6x \, dx + \int_{1}^{4} 9 \, dx \right] \]
4. **Evaluate each integral:**
\[ \int_{1}^{4}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe4498e0d-5915-421f-a7a7-3ad8a56e2358%2Fdc438922-b016-468b-8b92-bd682d0579f0%2F620d4zs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Topic: Calculating the Volume of Solid of Revolution**
**Problem Statement:**
Find the volume of the solid generated by rotating the region enclosed by \( y = x + 3 \), \( y = 0 \), \( x = 1 \), and \( x = 4 \) about the x-axis. Do not simplify your answer.
**Graphical Representation:**
The provided diagram illustrates the region enclosed by the following boundaries:
- The line \( y = x + 3 \), which is depicted in blue.
- The horizontal line \( y = 0 \) (the x-axis), shown in black.
- Two vertical lines representing \( x = 1 \) and \( x = 4 \), marked in red.
The shaded gray region indicates the area that will be revolved around the x-axis to form the solid.
**Explanation and Steps to Find the Volume:**
To find the volume of the solid formed by rotating the enclosed region around the x-axis, we use the disk method. The volume \( V \) is given by the integral:
\[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]
Here, the function \( f(x) = x + 3 \), and the limits of integration are from \( x = 1 \) to \( x = 4 \).
1. **Identify the function and limits:**
- \( f(x) = x + 3 \)
- Lower limit \( a = 1 \)
- Upper limit \( b = 4 \)
2. **Set up the integral:**
\[ V = \pi \int_{1}^{4} (x + 3)^2 \, dx \]
3. **Perform the integration:**
- Expand the integrand first:
\[ (x + 3)^2 = x^2 + 6x + 9 \]
- Integrate each term separately:
\[ V = \pi \int_{1}^{4} (x^2 + 6x + 9) \, dx \]
\[ V = \pi \left[ \int_{1}^{4} x^2 \, dx + \int_{1}^{4} 6x \, dx + \int_{1}^{4} 9 \, dx \right] \]
4. **Evaluate each integral:**
\[ \int_{1}^{4}
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