$y = \ln x$, region is delimited by $y = -1$, $y = 2$ and the $y$-axis, it rotates around the $y$-axis.
It's quite simple to solve by using disk integration but I can't get it right with shell integration, I want to understand what I'm doing wrong:
$x = e^{y}$
$y = -1 \rightarrow x = 1/e$
$y = 2 \rightarrow x = e^2$
$$ V = 2\pi * \int_{1/e}^{e^2} x\ \ln x \;\mathrm{d}x $$
Integrating that does not get me to the right answer which is $\pi/2(e^4 - e^{-2})$
You have missed what I have marked as $(1)$, and made a mistake in writing the shell integration for $(2)$.
For the integral, you need to consider the height of each cylindrical shell. $y=2$ is the upper bound, so the height is $2-\ln x$. The volume each infinitesimal cylindrical shell contributes will be $height*2\pi*radius*d(radius)$, in your case $(2-\ln x)2\pi x dx$. Integrating this, we get:
$$\int_{\frac{1}{e}}^{e^2} 2 \pi x (2-\log (x)) \, dx=\frac{\left(e^6-7\right) \pi }{2 e^2}$$
Now we need the inner cylinder (which is not really a function, it is a simple cylinder, so I will calculate it as a regular cylinder). The volume will be:
$$3\pi (e^{-1})^2$$
Summing the two volume, you get:
$$\frac{\left(e^6-1\right) \pi }{2 e^2}$$