Ch. 8 - Techniques of Integration

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 8 - Techniques of Integration

Problem 8.8.90b

Chapter 8, Problem 8.8.90b

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

Verified step by step guidance

Identify the region bounded by the curves: \(y = \frac{1}{\sqrt{x}}\), \(y = 0\), \(x = 0\), and \(x = 1\). This region lies in the first quadrant between \(x=0\) and \(x=1\) under the curve \(y = \frac{1}{\sqrt{x}}\).

Since the region is revolved about the x-axis, use the disk method to find the volume. The volume \(V\) is given by the integral \(V = \pi \int_{a}^{b} [f(x)]^{2} \, dx\), where \(f(x)\) is the function representing the radius of the disks.

Here, the radius of each disk is \(y = \frac{1}{\sqrt{x}}\). So, the volume integral becomes \(V = \pi \int_{0}^{1} \left( \frac{1}{\sqrt{x}} \right)^{2} \, dx\).

Simplify the integrand: \(\left( \frac{1}{\sqrt{x}} \right)^{2} = \frac{1}{x}\). So the integral is \(V = \pi \int_{0}^{1} \frac{1}{x} \, dx\).

Evaluate the integral \(\int_{0}^{1} \frac{1}{x} \, dx\). Note that this integral is improper at \(x=0\), so consider it as a limit: \(\lim_{t \to 0^{+}} \int_{t}^{1} \frac{1}{x} \, dx\). Set up this limit to analyze the behavior of 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 a solid. In this problem, the integral bounds are from 0 to 1, and the integral will be used to find the volume generated by revolving a region around an axis.

Volume of Solids of Revolution (Disk/Washer Method)

When a region is revolved around an axis, the volume of the resulting solid can be found using the disk or washer method. This involves integrating the area of circular cross-sections perpendicular to the axis of revolution, typically using the formula V = π∫[R(x)]² dx.

Understanding the Region Bounded by Curves

Identifying the region bounded by y = 1/√x, y = 0, x = 0, and x = 1 is crucial. This region lies in the first quadrant and defines the shape to be revolved. Correctly interpreting these boundaries ensures the integral is set up with proper limits and functions.

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Related Practice

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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Textbook Question

Consider the region bounded by the graphs of

y = arctan(x), y = 0, and x = 1.

b. Find the volume of the solid formed by revolving this region about the y-axis.

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 2 of x dx

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

Centroid:

Find the centroid of the region cut from the first quadrant by the curve

y = 1/√(x + 1) and the line x = 3.

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Textbook Question