$$x y = 1$$
$$y = 0$$
$$x = 1, x = 2$$
About $x = -1$
This question has been bugging me for a while and I can't seem to find a way to do it.
What I had done so far:
$$A_1(x) = \pi\left(1+\frac1y\right)^2$$
But then after that I don't know what else to do
Here is a plot to help visualize. Note that I shifted all the coordinates to the right by 1.
One way to simplify the problem is to translate the problem so that instead of rotating about $x=-1$, we can rotate about the $y$-axis. Which would mean that we integrate between $\frac{1}{x-1}$ (which is $\frac1x$ shifted to the right by 1) and $y=0$, and also shift our other bounds to the right by 1, making the bounds $x=2$ and $x=3$. I took the following formula right from the wiki page referenced above so that if you need clarification you can read the page.
Therefore,
$$V=2\pi \int_{2}^{3}x \,| \frac{1}{x-1}-0|\,dx$$ $$V=2\pi \int_{2}^{3}\frac{x}{x-1} \,dx$$ $$V = 2\pi (x+\log(x-1))\Big|_2^3$$
This simplifies to $$V = 2 \pi (1 + \log(2))$$