When the axis moves, the radii trade places AB & BC

Revolve about $y = k$ a region bounded by $y = f(x)$ above and $y = g(x)$ below. Each boundary contributes a distance from the axis:

$$\big|f(x) - k\big| \quad\text{and}\quad \big|g(x) - k\big|,$$

and the volume uses whichever is larger as $R$:

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

The error CA13 names is adjusting one of the two. Writing $R = f(x) - k$ while leaving $r = g(x)$ produces an integral that is well formed and evaluates cleanly, and there is nothing in the arithmetic to signal that half the setup describes a different solid.

The guard is symmetry of treatment: write both radii on the same line, in the same form, before doing anything else. If one has a $-k$ in it and the other does not, that is the defect, visible at a glance.

In 8.11 the axis was $y = 0$ and the region sat above it, so the upper boundary was also the farther one. That coincidence is what makes "outer" and "upper" feel like the same word, and it ends here.

Take $y = 2x$ above $y = x^{2}$ on $[0, 2]$, revolved about $y = 5$. Distances from the axis at $x = 1$:

1. To the line $y = 2$: a distance of $3$.
2. To the parabola $y = 1$: a distance of $4$.

The lower curve is farther from the axis, so $R = 5 - x^{2}$ and $r = 5 - 2x$. The boundaries have traded places, and

$$V = \pi\int_0^2 \Big(\big(5 - x^{2}\big)^{2} - \big(5 - 2x\big)^{2}\Big)\,dx = \pi\left[\frac{446}{15} - \frac{62}{3}\right] = \frac{136\pi}{15}.$$

Keeping the 8.11 assignment gives $-\frac{136\pi}{15}$, and a negative volume is the clearest possible signal that the swap was missed. The rule in one line: the boundary nearer the axis gives $r$, and the axis decides which that is.

One test catches both the half-adjustment and the missed swap, and it costs a single substitution.

Evaluate both radii at an endpoint and compare them with the picture. At $x = 0$ in the example, the region runs from $y = 0$ to $y = 0$: both boundaries are at the origin, five units from the axis. So both radii should be $5$, and $5 - x^{2}$ and $5 - 2x$ both give $5$ there. A half-adjusted setup with $r = 2x$ would give $0$, which would claim the region reaches the axis at $x = 0$.

Three checks that together leave very little room:

1. Both radii carry the axis. Same $k$, same form.
2. $R \geq r$ throughout. Test one interior point; if it fails, the boundaries swapped and were not swapped back.
3. Where the curves meet, $R = r$. The washer degenerates to a circle there, and the integrand is zero.

That last one is strong. At $x = 0$ and $x = 2$ the two curves coincide, so the integrand must vanish at both ends whatever the axis is, and any setup that does not is wrong before a single antiderivative is taken.

Revolving about $x = k$ is the same statement rotated, with all of 8.10's consequences:

1. Slices are horizontal, so the integral is in $dy$ with $y$-limits.
2. Both boundaries are solved for $x$ in terms of $y$, keeping branch restrictions.
3. Both radii are $|g(y) - k|$, and the farther one is $R$.
4. The swap still applies. An axis to the right of the region makes the left boundary the outer one.

This combination is where the most work is lost on the exam, because three separate things have to go right at once: the variable, the conversion of both boundaries, and the assignment of $R$ and $r$. Each has its own check, and running all three takes less time than one antiderivative.

A closing observation about how far this generalises. Every topic from 8.7 onward has been the same integral, $\int A\,dx$. The shape of $A$ changed from a square to a triangle to a circle to a ring, and the length feeding it changed from a gap between curves to a distance from an axis. Nothing else has changed in six topics.