A region bounded by f(x)=√x +4₁ y=0₁) Find the volume of the solid formed by revolving the region about the x-axis. 6 WA00 5 4- 3 AR 4-3-2-11 1 2 3 4 5 6 2324 -2+ 3+ t √x + 4, y = 0, x = -3, and x = 0 is shown below.
A region bounded by f(x)=√x +4₁ y=0₁) Find the volume of the solid formed by revolving the region about the x-axis. 6 WA00 5 4- 3 AR 4-3-2-11 1 2 3 4 5 6 2324 -2+ 3+ t √x + 4, y = 0, x = -3, and x = 0 is shown below.
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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### Multiple Choice Answers:
- \(\frac{15}{2} \pi \)
- \(\frac{17}{2} \pi \) (correct answer highlighted)
- \(11 \pi \)
- \(16 \pi \)
### Solution:
To find the volume of the solid formed by revolving this region about the x-axis, we use the disk method. The volume \( V \) is given by:
\[ V = \pi \int_{-3}^{0} [f(x)]^2 \, dx \]
Substitute \( f(x) = \sqrt{x + 4} \):
\[ V = \pi \int_{-3}^{0} (\sqrt{x + 4})^2 \, dx \]
\[ V = \pi \int_{-3}^{0} (x + 4) \, dx \]
Evaluate the integral:
\[ V = \pi \left[ \frac{x^2}{2} + 4x \right]_{-3}^{0} \]
\[ V = \pi \left[ \left(\frac{0^2}{2} + 4 \cdot 0\right) - \left(\frac{(-3)^2}{2} + 4 \cdot (-3)\right) \right] \]
\[ V = \pi \left[ 0 - \left(\frac{9}{2} - 12\right) \right] \]
\[ V = \pi \left[0 - \left](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb111f9a2-d607-44d0-9cf1-983102308b09%2F674f7bf1-386b-4956-b764-bb1398bb099c%2Fibbx24k_processed.png&w=3840&q=75)
Transcribed Image Text:# Calculus Problem: Finding the Volume of a Solid of Revolution
## Problem Statement:
A region bounded by the function \( f(x) = \sqrt{x + 4} \), the x-axis ( \( y = 0 \) ), the line \( x = -3 \), and the line \( x = 0 \) is shown below.
### Objective:
Find the volume of the solid formed by revolving this region about the x-axis.
### Graph:
The graph displays the function \( f(x) = \sqrt{x + 4} \) in green, with x ranging from -4 to 6 and y ranging from -4 to 6. The shaded region, located between \( x = -3 \) and \( x = 0 \), represents the area under the curve \( f(x) \) and above the x-axis.
### Multiple Choice Answers:
- \(\frac{15}{2} \pi \)
- \(\frac{17}{2} \pi \) (correct answer highlighted)
- \(11 \pi \)
- \(16 \pi \)
### Solution:
To find the volume of the solid formed by revolving this region about the x-axis, we use the disk method. The volume \( V \) is given by:
\[ V = \pi \int_{-3}^{0} [f(x)]^2 \, dx \]
Substitute \( f(x) = \sqrt{x + 4} \):
\[ V = \pi \int_{-3}^{0} (\sqrt{x + 4})^2 \, dx \]
\[ V = \pi \int_{-3}^{0} (x + 4) \, dx \]
Evaluate the integral:
\[ V = \pi \left[ \frac{x^2}{2} + 4x \right]_{-3}^{0} \]
\[ V = \pi \left[ \left(\frac{0^2}{2} + 4 \cdot 0\right) - \left(\frac{(-3)^2}{2} + 4 \cdot (-3)\right) \right] \]
\[ V = \pi \left[ 0 - \left(\frac{9}{2} - 12\right) \right] \]
\[ V = \pi \left[0 - \left
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