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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 of revolution is \(2 \pi\) cubic units.

Step by step solution

01

Setup the Integral

First, set up the integral for finding the volume. We know that \(f(x)=(x-1)^{-\frac{1}{4}}\), so when we square it, we get: $$ f(x)^2 = [(x-1)^{-\frac{1}{4}}]^2 = (x-1)^{-\frac{1}{2}} $$ So, our volume will be given by: $$ V = \int_{1}^{2} \pi (x - 1)^{-\frac{1}{2}}\, dx $$
02

Apply Substitution

Next, apply substitution to simplify the function inside the integral by letting $$ u = x - 1 $$ Then, $$ du = dx $$ So, the integral becomes: $$ V = \pi \int_{u(1)}^{u(2)} u^{-\frac{1}{2}}\, du $$ Now we need to find the new bounds of the integral after the substitution by plugging in the original bounds of x = 1 and x = 2 into the substitution expression: $$ u(1) = 1 - 1 = 0 $$ $$ u(2) = 2 - 1 = 1 $$ So, the new bounds are from 0 to 1. Thus, the integral is now: $$ V = \pi \int_{0}^{1} u^{-\frac{1}{2}}\, du $$
03

Evaluate the Integral

Now, solve the integral: $$ V = \pi \int_{0}^{1} u^{-\frac{1}{2}}\, du = \pi \left[ 2u^{\frac{1}{2}} \right]_{0}^{1} $$ $$ V = \pi (2(1)^{\frac{1}{2}} - 2(0)^{\frac{1}{2}}) = \pi(2-0) = 2 \pi $$ So, the volume of the solid of revolution is \(2 \pi\) cubic units.

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