The radius starts at the axis AB & BC

Revolve about the horizontal line $y = k$. The radius of the slice at $x$ is the vertical distance from that line to the boundary curve:

$$r(x) = \big|f(x) - k\big|,\qquad V = \pi\int_a^b \big(f(x) - k\big)^{2}\,dx.$$

The absolute value bars are for clarity rather than for computation, since the square removes the sign anyway. What they record is that $r$ is a length, so a "negative radius" is a signal that the subtraction was written in the order that happens not to match the picture, not that anything is wrong with the solid.

Getting the order right is still worth doing, because it makes the expression readable and because in 8.11 and 8.12 two radii have to be compared:

1. Region above the axis: $r = f(x) - k$.
2. Region below the axis: $r = k - f(x)$.
3. Either way, check at one interior $x$ that the number comes out positive.

For the region between $y = \sqrt{x}$ and $y = 2$ on $[0, 4]$ revolved about $y = 2$, the curve is below the axis, so $r = 2 - \sqrt{x}$, and at $x = 1$ that is $1$, which matches the picture.

Continuing that example,

$$V = \pi\int_0^4 \big(2 - \sqrt{x}\big)^{2}\,dx = \pi\int_0^4 \big(4 - 4\sqrt{x} + x\big)\,dx = \pi\left[16 - \frac{64}{3} + 8\right] = \frac{8\pi}{3}.$$

Three ways this goes wrong, and they are worth separating because they have different causes:

1. Ignoring the shift: $r = \sqrt{x}$ gives $\pi\int_0^4 x\,dx = 8\pi$, which is the answer to the 8.9 problem about the $x$-axis. It is the reflex from the previous topic, applied where it does not hold.
2. Shifting the wrong way: $r = \sqrt{x} + 2$ gives $\frac{136\pi}{3}$. Adding when the situation calls for subtracting puts the axis on the far side of the curve, describing a solid several times too large.
3. Dropping $\pi$: $\frac{8}{3}$, which is CA14 unchanged from 8.9.

The cross term deserves its own mention. $\big(2 - \sqrt{x}\big)^{2}$ is $4 - 4\sqrt{x} + x$, and expanding it as $4 - x$ loses the middle term. Since the shift is exactly what creates a two-term radius, this topic is where that expansion error becomes common rather than occasional.

Revolving about $x = k$ makes the slices horizontal, because slices are always perpendicular to the axis of revolution. So the thickness is $dy$, the limits are $y$-values, and the boundary has to be solved for $x$:

$$V = \pi\int_c^d \big(g(y) - k\big)^{2}\,dy.$$

This is the most reliable place to lose a problem, because the radius can be handled perfectly while the integral is still set up in the wrong variable. The structural check from 8.5 applies without change: a $dy$ integral contains no $x$, in the integrand or in the limits.

1. Identify the axis and its direction. $x = k$ is vertical, so slices are horizontal.
2. Solve every boundary for $x$ in terms of $y$, keeping the branch restriction.
3. Convert the limits to the $y$-coordinates of the region's lowest and highest points.
4. Write $r = |g(y) - k|$ and square it.

Revolving $y = x^{2}$ on $[0, 3]$ about $x = 3$, for instance, becomes $\pi\int_0^9 \big(3 - \sqrt{y}\big)^{2}\,dy$: the curve solved for $x$, the distance measured from $x = 3$, and the limits taken from $y$.

One habit catches every error in this topic, and it takes about ten seconds.

Evaluate the radius at one interior point and compare it with the picture. Pick a convenient value, compute $r$, and ask whether that is really how far the boundary sits from the axis there.

In the running example at $x = 1$: the curve is at $y = 1$, the axis is at $y = 2$, so the radius should be $1$. The correct expression $2 - \sqrt{x}$ gives $1$; the reflex $\sqrt{x}$ gives $1$ as well, which is why this particular check needs a second point. At $x = 4$ the curve meets the axis, so the radius must be $0$: there $2 - \sqrt{x}$ gives $0$ and $\sqrt{x}$ gives $2$. The second point separates them.

That generalises into the rule worth keeping: test where the region touches the axis, since the radius has to vanish there. A radius expression that does not go to zero at a point of contact is measuring from the wrong line, and it will fail this check no matter how plausible it looks.