Find the volume using the disk method

Question

Let $R$ be the region bounded by the following curves. Let $S$ be the solid generated when $R$ is revolved about the $x$-axis. Find the volume using the disk method.

$$\begin{align*} &y=\sqrt{\ln x}\\ &y=\sqrt{\ln x^2}\\ &y=1 \end{align*}$$

I cannot seem to get the right answer for this. I am getting the result of just $\pi$ and that is incorrect. I first found where the two natural log equations intersect and that is at $1$. So that would be my lower limit. I found my upper limit by setting the first equation equal to the last equation ($= 1$). That gave me my upper limit of $e$. I removed the $\pi$ outside of the integration and then squared both of the natural log equations. Therefore, I would be integrating $\ln(x^2)-\ln(x)$ because the $ln(x^2)$ equation is above the other one. After integrating, I got the answer of $1$. I then multiplied this by $\pi$ to get the answer of $\pi$. However, the website is saying this is incorrect. What did i do wrong?

Answer 1

The region $R$ has three sides. Between $1$ and $\sqrt{e}$, the disks have outer radius given by $\sqrt{\ln(x^2)}$ and inner radius $\sqrt{\ln x}$. However between $\sqrt{e}$ and $e$, the disks have outer radius given by $1$ and inner radius given by $\sqrt{\ln x}$.

See diagram:

Answer 2

With the disk method you need two integrals, one for the part of the region to the left of the point of intersection of $y=1$ and $y=\sqrt{\ln x^2}$ and one for the part to the right. That intersection occurs where $1=\ln x^2=2\ln x$, i.e., at $x=\sqrt{e}$. On the interval from $x=1$ to $x=\sqrt{e}$ the region $R$ lies between the curves $y=\sqrt{\ln x^2}$ (at the top) and $y=\sqrt{\ln x}$ (at the bottom); on the interval from $x=\sqrt{e}$ to $x=e$, however, it lies between $y=1$ (at the top) and $y=\sqrt{\ln x}$ (at the bottom).

The disk method is a poor choice here: the shell method sets up more neatly, requiring only a single integral.