Problem 51 - Calculus Early Transcendentals

Question

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.

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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. $$ V = \int_1^2 \pi [(x-1)^{-1/4}]^2 dx $$

Step 3: Simplify the integral

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

Step 4: 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: $$ V = \pi \int_0^1 u^{-1/2} du = \pi \left[ 2u^{1/2} \right]_0^1 = \pi \left[ 2(1)^{1/2} - 2(0)^{1/2} \right] = 2\pi $$

Step 5: 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.