I want to make sure I'm understanding this correctly. If a continuous, nonnegative function $f(x)$ on $[a,b]$ was revolved about some axis, where it grows by $x$ in $\Delta x$ intervals, then the rate of change of volume is the area of the base of the cylinder. So then volume is the area times its height, then $$dV=\pi((r_o)^2-(r_i)^2)dx$$
If instead, the cylinder's height grows by $y$ in $\Delta y$ intervals then $$dV=\pi((f^{-1}_o(y))^2-(f^{-1}_i(y))^2)dy$$ Where the radii is $x$ in terms of $y$.
Now if the radius grows by $x$ in $\Delta x$ intervals, then the rate of change of volume is no longer the area of the base, but the area of the lateral side of the cylinder. Then $$dV=2\pi x f(x) dx$$ If instead, the radius grows by $y$ in $\Delta y$ intervals, then $$dV = 2\pi f^{-1}(y)y dy$$
Am I describing this correctly?