Volume with Washer Method: Revolving Around the x- or y-Axis AB & BC

A washer is a disc with a smaller disc removed, so its area is $\pi(R^{2} - r^{2})$ and the volume is $\pi\int_a^b (R^{2} - r^{2})\,dx$, with $R$ the distance from the axis to the far boundary and $r$ the distance to the near one. That expression is two disc integrals subtracted, and it is never $\pi(R - r)^{2}$: for $R = 5$ and $r = 3$ the two give $16\pi$ and $4\pi$. $R - r$ is the width of the ring, and a thin ring at a large radius has far more area than a small disc, because it is stretched around a long circumference.

The outer radius reaches whichever boundary is farther from the axis, decided by testing a point inside the interval rather than read off the equations. For $y = 2x$ and $y = x^{2}$ on $[0, 2]$ about the $x$-axis, $R = 2x$ and $r = x^{2}$ give $\frac{64\pi}{15}$; dropping the inner radius gives $\frac{32\pi}{3}$ and squaring the difference gives $\frac{16\pi}{15}$. A region that touches the axis has $r = 0$ there and needs a disc, a region that stops short needs a washer, and a region that does both needs a split. About the $y$-axis both boundaries must be solved for $x$, so the conversion has two chances to fail rather than one.