Let $S$ be region enclosed by $y= 1/x$, $y=1$ and $y=2$. Goal is to find the volume of $S$ when it is revolved about $x= -1$
try:
Best to use washer method wrt y coordinate. A volume element is
$$ dV = ( \pi (1 + 1/y)^2 - \pi (-1)^2 ) dy $$ Thus,
$$ V = \int_1^2 (1 + 2/y + 1/y^2 - 1 ) dy = \int_1^2 (2/y + 1/y^2) dy $$
is this a correct setting for the required volume?
The shaded area below is the region $S$.
Let us write $f(y)=\frac{1}{y}$. The outer radius is $f(y)+1$ and the inner radius is $1$. Thus, by using the washer method, $$dV=\pi\bigg(\big[f(y)+1\big]^2-1^2\bigg)dy$$ so that the volume is given by $$V=\int_{1}^2 dV=\int_{1}^2\pi\bigg(\big[f(y)+1\big]^2-1^2\bigg)dy=\pi\int_1^2 \left(\frac{2}{y} + \frac{1}{y^2}\right) dy.$$ Your only mistake is you introduced $(-1)^2$ and that you don't have $\pi$ in your integral.