A solid of revolution is generated by rotating the region between the x-axis and the graphs of f(x) = √√x+1, x = 3, and x=8 about the x-axis. Using the disk method, what is the volume of the solid? Enter only the value of the volume and do not include units in your answer. Submit your answer in terms of #, if necessary.
A solid of revolution is generated by rotating the region between the x-axis and the graphs of f(x) = √√x+1, x = 3, and x=8 about the x-axis. Using the disk method, what is the volume of the solid? Enter only the value of the volume and do not include units in your answer. Submit your answer in terms of #, if necessary.
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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![**Volume of a Solid of Revolution Using the Disk Method**
**Problem Statement:**
A solid of revolution is generated by rotating the region between the x-axis and the graphs of \( f(x) = \sqrt{x} + 1 \), \( x = 3 \), and \( x = 8 \) about the x-axis. Using the disk method, what is the volume of the solid?
Enter only the value of the volume and do not include units in your answer. Submit your answer in terms of \(\pi\), if necessary.
**Provide your answer below:**
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**Feedback Button:**
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**Explanation:**
The above problem involves finding the volume of a solid created by rotating a given region around the x-axis. The function provided is \( f(x) = \sqrt{x} + 1 \). We are given the limits \( x = 3 \) and \( x = 8 \).
To solve this problem using the disk method, we integrate the area of the circular cross-sections of the solid along the axis of rotation. In the disk method, these cross-sections are circles with radius \( f(x) \).
The volume \( V \) of the solid of revolution is calculated using the integral:
\[ V = \pi \int_{3}^{8} [f(x)]^2 \, dx \]
Substituting \( f(x) = \sqrt{x} + 1 \):
\[ V = \pi \int_{3}^{8} (\sqrt{x} + 1)^2 \, dx \]
We then need to expand the integrand and evaluate the integral:
\[ (\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1 \]
So the integral becomes:
\[ V = \pi \int_{3}^{8} (x + 2\sqrt{x} + 1) \, dx \]
Evaluating this integral will give the volume of the solid.
**Further Steps:**
1. Integrate \( \pi \int_{3}^{8} x \, dx \)
2. Integrate \( \pi \int_{3}^{8} 2\sqrt{x} \, dx \)
3. Integrate \( \pi \int_{3}^{8} 1 \, dx \)
4. Sum the results and simplify to find the volume.
Finally, make sure to submit your](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa79c18d0-6a0e-4a38-9496-04f82bd62d69%2Fd6c114e1-be71-4b1d-8635-3d56fd63e925%2F0peq2kj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Volume of a Solid of Revolution Using the Disk Method**
**Problem Statement:**
A solid of revolution is generated by rotating the region between the x-axis and the graphs of \( f(x) = \sqrt{x} + 1 \), \( x = 3 \), and \( x = 8 \) about the x-axis. Using the disk method, what is the volume of the solid?
Enter only the value of the volume and do not include units in your answer. Submit your answer in terms of \(\pi\), if necessary.
**Provide your answer below:**
[Input Field]
**Feedback Button:**
[FEEDBACK Button]
**Explanation:**
The above problem involves finding the volume of a solid created by rotating a given region around the x-axis. The function provided is \( f(x) = \sqrt{x} + 1 \). We are given the limits \( x = 3 \) and \( x = 8 \).
To solve this problem using the disk method, we integrate the area of the circular cross-sections of the solid along the axis of rotation. In the disk method, these cross-sections are circles with radius \( f(x) \).
The volume \( V \) of the solid of revolution is calculated using the integral:
\[ V = \pi \int_{3}^{8} [f(x)]^2 \, dx \]
Substituting \( f(x) = \sqrt{x} + 1 \):
\[ V = \pi \int_{3}^{8} (\sqrt{x} + 1)^2 \, dx \]
We then need to expand the integrand and evaluate the integral:
\[ (\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1 \]
So the integral becomes:
\[ V = \pi \int_{3}^{8} (x + 2\sqrt{x} + 1) \, dx \]
Evaluating this integral will give the volume of the solid.
**Further Steps:**
1. Integrate \( \pi \int_{3}^{8} x \, dx \)
2. Integrate \( \pi \int_{3}^{8} 2\sqrt{x} \, dx \)
3. Integrate \( \pi \int_{3}^{8} 1 \, dx \)
4. Sum the results and simplify to find the volume.
Finally, make sure to submit your
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