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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![### Calculus Problem: Region and Solid of Revolution
**Problem Statement:**
Let \( R \) be the region bounded by the curve \( y = \frac{1}{x} \), the vertical line \( x = 1 \), and the x-axis on the interval \([1, +\infty)\). Let \( S \) be the solid generated by rotating the region \( R \) about the x-axis.
**Tasks:**
1. **Find the volume of \( S \) if it's finite or show that it's infinite.**
2. **Show that the surface area of the solid \( S \) is infinite.**
**Instructions and Solutions:**
1. **Volume Calculation:**
To find the volume of the solid \( S \), we use the method of disks or washers. The volume \( V \) of the solid obtained by rotating the region \( R \) about the x-axis is given by the integral:
\[
V = \pi \int_{1}^{\infty} \left( \frac{1}{x} \right)^2 dx
\]
Simplifying inside the integral:
\[
V = \pi \int_{1}^{\infty} \frac{1}{x^2} dx
\]
Solving this integral:
\[
\int \frac{1}{x^2} dx = -\frac{1}{x}
\]
Thus,
\[
V = \pi \left[ -\frac{1}{x} \right]_{1}^{\infty} = \pi \left( 0 + 1 \right) = \pi
\]
Therefore, the volume \( V \) is finite and equals \( \pi \).
2. **Surface Area Calculation:**
The surface area \( A \) of the solid of revolution is given by:
\[
A = 2\pi \int_{1}^{\infty} \frac{1}{x} \sqrt{1 + \left( \frac{d}{dx} \left( \frac{1}{x} \right) \right)^2 } dx
\]
Simplifying inside the integral:
\[
\frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0a0c2eed-fc4c-4fe4-9822-86e8a241e03b%2F94138aa0-0b67-4d99-95f0-7001970e8c00%2Fjkvx4yc_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Calculus Problem: Region and Solid of Revolution
**Problem Statement:**
Let \( R \) be the region bounded by the curve \( y = \frac{1}{x} \), the vertical line \( x = 1 \), and the x-axis on the interval \([1, +\infty)\). Let \( S \) be the solid generated by rotating the region \( R \) about the x-axis.
**Tasks:**
1. **Find the volume of \( S \) if it's finite or show that it's infinite.**
2. **Show that the surface area of the solid \( S \) is infinite.**
**Instructions and Solutions:**
1. **Volume Calculation:**
To find the volume of the solid \( S \), we use the method of disks or washers. The volume \( V \) of the solid obtained by rotating the region \( R \) about the x-axis is given by the integral:
\[
V = \pi \int_{1}^{\infty} \left( \frac{1}{x} \right)^2 dx
\]
Simplifying inside the integral:
\[
V = \pi \int_{1}^{\infty} \frac{1}{x^2} dx
\]
Solving this integral:
\[
\int \frac{1}{x^2} dx = -\frac{1}{x}
\]
Thus,
\[
V = \pi \left[ -\frac{1}{x} \right]_{1}^{\infty} = \pi \left( 0 + 1 \right) = \pi
\]
Therefore, the volume \( V \) is finite and equals \( \pi \).
2. **Surface Area Calculation:**
The surface area \( A \) of the solid of revolution is given by:
\[
A = 2\pi \int_{1}^{\infty} \frac{1}{x} \sqrt{1 + \left( \frac{d}{dx} \left( \frac{1}{x} \right) \right)^2 } dx
\]
Simplifying inside the integral:
\[
\frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1
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