Skip to main content
Ch. 8 - Techniques of Integration
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook
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
1
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.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
3m
Was this helpful?

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.
Recommended video:
05:43
Definition of the Definite Integral

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.
Recommended video:
04:48
Finding Volume Using Disks

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.
Recommended video:
05:06
Finding Area When Bounds Are Not Given