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Chapter 7: Problem 51

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=(x-1)^{-1 / 4}\) and the \(x\) -axis on the interval (1,2] is revolved about the \(x\) -axis.

Short Answer

Expert verified
Answer: The volume of the solid is \(2\pi\) cubic units.

Step by step solution

01

Define the function

The function \(f(x)=(x-1)^{-1/4}\) is the curve that bounds the region we are considering. We will use this function as the radius of the disks in our calculation of the volume.
02

Set up the integral formula

We'll use the disk method to find the volume of the solid of revolution described by revolving the region bounded by \(f(x)\) and the x-axis on the interval (1, 2] around the x-axis. The volume of a single thin disk with radius \(f(x)\) and thickness \(dx\) is given by \(V_{disk} = \pi [f(x)]^2dx\). To find the total volume, we will sum (integrate) these individual disk volumes along the x-axis from 1 to 2. $$ V = \int_1^2 \pi [(x-1)^{-1/4}]^2 dx $$
03

Simplify the integral

To simplify the integral, we will square the radius function inside the square brackets: $$ V = \int_1^2 \pi [(x-1)^{-1/2}] dx $$
04

Evaluate the integral

Now let's evaluate the integral: $$ V = \pi \int_1^2 (x-1)^{-1/2} dx $$ To evaluate the integral, we use a u-substitution: let \(u = x-1\), then \(du = dx\). When \(x = 1\), \(u = 0\), and when \(x = 2\), \(u = 1\). Now we can rewrite and evaluate the integral in terms of u: $$ V = \pi \int_0^1 u^{-1/2} du = \pi \left[ 2u^{1/2} \right]_0^1 = \pi \left[ 2(1)^{1/2} - 2(0)^{1/2} \right] = 2\pi $$
05

State the final answer

The volume of the solid generated by revolving the given region around the x-axis on the interval (1, 2] is \(2\pi\) cubic units.

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