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Chapter 7: Problem 56

Find the volume of the described solid of revolution or state that it does not exist. The region bounded by \(f(x)=-\ln x\) and the \(x\) -axis on the interval (0,1] is revolved about the \(x\) -axis.

Short Answer

Expert verified
In this problem, we revolved the region bounded by the function \(f(x)=-\ln x\) and the \(x\)-axis on the interval \((0,1]\) around the \(x\)-axis. By setting up an integral using the method of disks/washers, substituting the function into the integral expression, and solving the integral, we found out that the volume of the solid of revolution is negative infinity. This implies that the volume does not exist since the region described is unbounded on the given interval.

Step by step solution

01

1. Setting up the integral for the volume of a solid of revolution

First, let's set up the integral for the volume of the solid. In this problem, since we are revolving around the \(x\)-axis, the elemental volume of a thin disk with thickness \(\Delta x\) is given by \(V_{disk} = \pi r^2 \Delta x\), where here, the radius \(r\) is defined by the function \(-\ln x\). Now let's consider the integral expression for the volume of the entire solid: $$V=\int_a^b \pi r^2 dx$$ Here, \(a\) and \(b\) are the limits of integration, which, in this case, are given as (0, 1].
02

2. Substituting the function into the integral

Since \(r=-\ln x\), we need to substitute this into the integral expression for the solid's volume: $$V=\int_a^b \pi (-\ln x)^2 dx$$ Now, we'll plug in the correct limits of integration. $$V=\int_0^1 \pi (-\ln x)^2 dx$$
03

3. Solving the integral

Now we need to solve the integral. Let's start by making a substitution: let \(u=-\ln x\). Then, we have that \(\frac{du}{dx}=\frac{1}{x}\) or \(\frac{dx}{du}=-u\). We also need to transform the limits of integration: when \(x=0\), \(u=\infty\) and when \(x=1\), \(u=0\). So the integral becomes: $$V=\int_\infty^0 \pi u^2 (-u) du$$ We will now change the limits again and switch the limits of integration to make the integral easier to work with: $$V=-\int_0^\infty \pi u^3 du$$ Now we can solve this integral by using the power rule: $$V=-\pi \int_0^\infty u^3 du = -\pi \left[\frac{1}{4}u^4\right]_0^\infty$$
04

4. Evaluating the integral and interpreting the result

Now we need to evaluate the limits: $$V=-\pi \left(\frac{1}{4}\infty^4 - \frac{1}{4}(0)^4\right)= -\infty$$ The result of the integral is negative infinity, which implies that the volume of the described solid of revolution does not exist. This is due to the fact that the region bounded by the function \(f(x)=-\ln x\) and the \(x\)-axis is unbounded on the interval \((0,1]\).

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