Textbook Question
88. The region in Exercise 87 is revolved about the x-axis to generate a solid.
b. Show that the inner and outer surfaces of the solid have infinite area.
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Ch. 8 - Techniques of Integration
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 8, Problem 8.8.89b
89. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / x², y = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (i) about the x-axis.
Verified step by step guidance1
Identify the region to be revolved: it is bounded by the curve \(y = \frac{1}{x^2}\), the x-axis (\(y=0\)), and the vertical line \(x=1\), in the first quadrant. Since the region is in the first quadrant, \(x\) ranges from 1 to infinity.
Set up the volume integral using the disk method because the region is revolved around the x-axis. The radius of each disk is the distance from the x-axis to the curve, which is \(r(x) = \frac{1}{x^2}\).
Write the volume integral as \(V = \pi \int_{1}^{\infty} \left(r(x)\right)^2 \, dx = \pi \int_{1}^{\infty} \left(\frac{1}{x^2}\right)^2 \, dx\).
Simplify the integrand: \(\left(\frac{1}{x^2}\right)^2 = \frac{1}{x^4}\), so the integral becomes \(V = \pi \int_{1}^{\infty} \frac{1}{x^4} \, dx\).
Evaluate the improper integral by finding the antiderivative of \(x^{-4}\), then take the limit as the upper bound approaches infinity to determine the volume.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Definite Integrals for Area and Volume
Definite integrals calculate the accumulation of quantities, such as area under a curve or volume of solids. In volume problems, integrals sum infinitesimal cross-sectional areas along an axis to find total volume. Setting correct limits and integrand expressions is essential.
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05:43
Definition of the Definite Integral
Method of Disks/Washers for Volumes of Revolution
The disk/washer method finds volumes by slicing the solid perpendicular to the axis of revolution. Each slice forms a disk or washer whose area is integrated along the axis. For revolution about the x-axis, the radius is the function value, and volume is π∫[radius]^2 dx.
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04:48Finding Volume Using Disks
Understanding the Region and Boundaries
Identifying the region bounded by y = 1/x², y = 0, and x = 1 in the first quadrant is crucial. This defines the limits of integration and the shape being revolved. Recognizing that the region extends from x=1 to infinity helps set proper integral bounds for volume calculation.
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07:45Area of Polar Regions
Related Practice
Textbook Question
90. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / √x, y = 0, x = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (i) about the x-axis
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Textbook Question
Finding area
Find the area of the region enclosed by the curve y = x sin(x) and the x-axis (see the accompanying figure) for:
b. π ≤ x ≤ 2π.
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Textbook Question
Finding volume: Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes and the curve y = cos(x), 0 ≤ x ≤ π/2, about
b. The line x = π/2.
Textbook Question
In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from 0 to 3 of 1/√(x + 1) dx
Textbook Question
In Exercises 11–22, estimate the minimum number of subintervals needed to approximate the integrals with an error of magnitude less than 10^-4 by (b) Simpson’s Rule. (The integrals in Exercises 11–18 are the integrals from Exercises 1–8.)
∫ from -1 to 1 of (x² + 1) dx