Difference of squares, not square of a difference AB & BC

A washer is a disc with a smaller disc removed. Its area is the difference of the two areas:

$$A = \pi R^{2} - \pi r^{2} = \pi\big(R^{2} - r^{2}\big),$$

so the volume of the solid is

$$V = \pi\int_a^b \Big(\big(R(x)\big)^{2} - \big(r(x)\big)^{2}\Big)\,dx.$$

$R$ is the distance from the axis of revolution to the far boundary of the region and $r$ the distance to the near one. Both are radii in the sense of 8.9 and 8.10: distances from the axis, one per boundary.

The structure is worth naming. This is not a new method; it is two disc integrals subtracted. The solid swept by the whole region under the outer boundary, minus the solid swept by the region under the inner one, is the solid with the hole. Everything else follows from that sentence.

The single most common error in this topic writes $\pi\big(R - r\big)^{2}$, treating the hole as a correction to the radius. Put numbers in it:

1. $R = 5$, $r = 3$: $R^{2} - r^{2} = 16$, while $(R - r)^{2} = 4$.
2. $R = 5$, $r = 4$: $R^{2} - r^{2} = 9$, while $(R - r)^{2} = 1$.
3. $R = 5$, $r = 1$: $R^{2} - r^{2} = 24$, while $(R - r)^{2} = 16$.

They never agree unless $r = 0$, which is exactly the case where the washer collapses into a disc. And the geometry explains why: $R - r$ is the width of the ring, and a ring of width $2$ at radius $5$ has far more area than a disc of radius $2$, because it is stretched around a large circumference.

The algebra says the same thing. $(R - r)^{2} = R^{2} - 2Rr + r^{2}$, which is not $R^{2} - r^{2}$ under any circumstances but $r = 0$. This is the same false identity that appeared in 8.7 on squared side lengths and in 8.10 on shifted radii, and it is worth recognising as one recurring error rather than three.

The outer radius reaches the boundary farther from the axis of revolution. About the $x$-axis with a region above it, that is the upper curve.

1. Find the intersections to fix the interval, as in 8.4.
2. Test a point inside to see which curve is farther from the axis.
3. Assign $R$ and $r$ and keep them for the whole subinterval.

For $y = 2x$ and $y = x^{2}$ on $[0, 2]$ about the $x$-axis, testing $x = 1$ gives $2$ and $1$, so $R = 2x$ and $r = x^{2}$, and

$$V = \pi\int_0^2 \big(4x^{2} - x^{4}\big)\,dx = \pi\left[\frac{32}{3} - \frac{32}{5}\right] = \frac{64\pi}{15}.$$

Two failures to watch. Dropping the inner radius gives $\pi\int_0^2 4x^{2}\,dx = \frac{32\pi}{3}$, the solid with its hole filled in. Writing $\pi\int_0^2 (2x - x^{2})^{2}\,dx$ gives $\frac{16\pi}{15}$, which is the square-of-the-difference error and, notably, comes out smaller than the truth rather than larger.

Note also that "farther from the axis" and "on top" are the same thing only while the region sits above a horizontal axis. Topic 8.12 moves the axis and the two can come apart.

The decision is 8.9's condition, stated once more from the other side:

1. The region touches the axis of revolution: $r = 0$ on that stretch, so the slice is a disc.
2. The region stops short of the axis: $r > 0$, so the slice is a washer.

A region can do both on different parts of its interval, and then the integral splits at the changeover, exactly as an area splits at a crossing in 8.6.

Revolving about the $y$-axis is the same statement with the letters exchanged: solve both boundaries for $x$ in terms of $y$, take the far one as $R$ and the near one as $r$, and integrate in $dy$ between $y$-limits. The structural check from 8.5 still applies, and here it has two chances to fail, since two boundaries need converting rather than one.

Finally, $\pi$. It multiplies the whole integrand, so it applies to both radii at once. Writing $\pi R^{2} - r^{2}$, with the constant attached to only one term, is a real and easily missed slip and it is dimensionally inconsistent as well.