What is the volume of the solid generated by revolving about the $y$-axis, the area bounded by the parabola $y=(x-1)^2-1, y=2$, and the $y$-axis?
I'm confused with the outer and inner radius. I am a little bit skeptical in using the formula of the volume of solid revolution. I don't know where to substitute those values in the formula.
In order to gain some intuition about this problem, have a look at the following figure (produced by Wolfram alpha).
What is interesting about this problem is that the area enclosed between the two functions, namely $f_1(x) = (x-1)^2 - 1$ and $f_2(x) =2$, intersects with the $y$ axis which is the axis of rotation.
We therefore need to revolve the surface enclosed between $f_1$ and $f_2$ and $x\geq 0$ with area
$$A = \int_0^{1+\sqrt{3}}(f_2(x) - f_1(x))\mathrm{d}x$$
This can be easily seen to be
$$V = \int_0^{1+\sqrt{3}}2\pi (f_2(x) - f_1(x))x \mathrm{d}x $$
If, on the other hand, we assume that the revolution creates a cavity in the solid body, then we need to subtract the following volume
$$ \Delta V = \int_0^{1+\sqrt{3}}2\pi (f_2(x) - f_1(-x))x \mathrm{d}x $$