Quick questions on Volumes of revolution: discs about the x-axis and y-axis

Consider a region bounded by $y = f(x)$, the x-axis, and the vertical lines $x = a$ and $x = b$ (with $f(x) \ge 0$ on $[a, b]$).

The disc method is best seen as a short film: a flat region spins about an axis, sweeps out a solid, and that solid is then sliced back into discs to add up. The four panels build exactly that picture for the region under $y = \sqrt{x}$ from $x = 0$ to $x = 4$, rotated about the x-axis.

For a region with $x = g(y)$ on $[c, d]$ (with $g(y) \ge 0$), rotating about the y-axis gives discs of radius $g(y)$ and thickness $dy$, so

If the region is bounded above by $y = f(x)$ and below by $y = h(x)$ (with $0 \le h(x) \le f(x)$) and rotated about the x-axis, each cross section is an annulus, a washer, whose area is the big circle minus the hole, $\pi R^2 - \pi r^2$:

Rotation about the x-axis uses vertical strips ($dx$); rotation about the y-axis uses horizontal strips ($dy$).

If you switch from $dx$ to $dy$, the limits change to the y-range of the region.

Outer minus inner, both squared. Reversing the order or forgetting to subtract gives nonsense.