Volume of revolution of the area between $\sec(x-1)$ and $\ln(x)$ for $1 \leq x \leq 2$

Question

I am trying to compute the volume of the solid of revolution given by the rotation of the area between $\sec(x-1)$ and $\ln(x)$ for $1 \leq x \leq 2$, rotated around the $x$-axis. I have tried using the disk method, $$V = \pi \int_1^2 (\sec(x-1)-\ln(x))^2 \, dx,$$ as well as the shell method, $$V = \pi \int_1^2 x(\sec(x-1) - \ln(x))\, dx,$$ but none worked. Mathematica didn't compute any of these integrals. I've tried by hand but I run into other more horrible integrals as $$\int_1^2 \sec(x-1) \ln(x) \, dx.$$

Answer 1

The wrong formula was used for the Disk Method, it should be the integral from $1$ to $2$ of $\pi(\sec^2(x-1)-\ln^2 x)$. (The Shell Method was also incorrectly set up, that integral represents volume when we rotate about the $y$-axis.)

An antiderivative of $\sec^2(x-1)$ is immediate, it is $\tan(x-1)$. To integrate $\ln^2 x$, integrate by parts, letting $u=\ln^2 x$ and $dv=1\,dx$. Then we will need to integrate by parts one more time.