A slice that is a circle AB & BC

Revolve the region under $y = f(x)$, above the $x$-axis, from $x = a$ to $x = b$, about the $x$-axis. A slice perpendicular to the axis is a circle of radius $f(x)$, so $A = \pi\big(f(x)\big)^{2}$ and

$$V = \pi\int_a^b \big(f(x)\big)^{2}\,dx.$$

Nothing here is new except the shape. It is still $\int A\,dx$; the cross section is simply a circle, and the radius is a distance measured from the axis of revolution to the boundary curve.

For $y = \sqrt{x}$ on $[0, 4]$ about the $x$-axis,

$$V = \pi\int_0^4 \big(\sqrt{x}\big)^{2}\,dx = \pi\int_0^4 x\,dx = 8\pi.$$

Revolving about the $y$-axis is the same statement with the letters exchanged: solve the boundary for $x$ in terms of $y$, and integrate $\pi\big(g(y)\big)^{2}\,dy$ between $y$-limits. That is 8.5's conversion, reused, and the same structural check applies: a $dy$ volume integral contains no $x$.

Three arithmetic failures account for nearly every wrong disc answer, and all three are CA14. Against the true $8\pi$ for the example above:

1. Omitting $\pi$. $\int_0^4 x\,dx = 8$. The number is otherwise correct and the answer is not a volume of revolution at all.
2. Omitting the square. $\pi\int_0^4 \sqrt{x}\,dx = \frac{16\pi}{3}$. This integrates the radius rather than the area.
3. Squaring after integrating. $\pi\left(\int_0^4 \sqrt{x}\,dx\right)^{2} = \frac{256\pi}{9}$. Squaring is part of forming the integrand, so it happens before the integral sign, not after it.

The third is the one worth dwelling on, because it looks like bookkeeping and is a real misunderstanding: $\int (f)^{2} \neq \left(\int f\right)^{2}$. Each slice's area is $\pi r^{2}$ for that slice's own radius, and adding areas is not the same as squaring a sum of radii.

A related error appears whenever the radius is a difference. Then $r^{2} = (A - B)^{2} = A^{2} - 2AB + B^{2}$, and expanding it as $A^{2} - B^{2}$ drops the cross term. That is 8.7's false expansion returning, and it becomes the central error of 8.11.

Writing $r = f(x)$ is correct when the axis of revolution is the $x$-axis, because the distance from $y = 0$ up to the curve is exactly $f(x)$. It is a special case, not a definition.

The general statement is that $r$ is the distance from the axis of revolution to the boundary being revolved. Since a distance is never negative, $r$ is written so that it comes out positive on the whole interval. Topic 8.10 takes up the case where the axis is a line like $y = 3$ and the distance becomes $|f(x) - 3|$.

Two consequences hold already:

1. A region below the axis gives the same solid as its reflection. Revolving $y = -\sqrt{x}$ about the $x$-axis produces exactly the solid $y = \sqrt{x}$ does, because the radius is a distance and $\left(-\sqrt{x}\right)^{2} = x$. The squaring handles the sign for you.
2. The variable of integration is set by the axis. Revolving about the $x$-axis makes slices perpendicular to it, so the thickness is $dx$ and the limits are $x$-values. Revolving about the $y$-axis makes them $dy$ and $y$-values.

A disc is a solid circle, so it assumes the solid has no hole along that slice. That is true exactly when the region being revolved touches the axis of revolution.

Check it before writing $\pi r^{2}$:

1. Look at the boundary nearest the axis. If it is the axis itself, discs are correct.
2. If some other curve is nearer, revolving leaves a hole and the slice is a washer, which is 8.11.

The region under $y = \sqrt{x}$ from $0$ to $4$ sits on the $x$-axis, so discs are right. The region between $y = \sqrt{x}$ and $y = 1$ from $x = 1$ to $x = 4$ does not touch the $x$-axis anywhere, so revolving it about that axis makes a solid with a cylindrical hole of radius $1$ through it, and $\pi\int (\sqrt{x})^{2}\,dx$ would count the hole as part of the solid.

That error is CA11, and it is not detectable in the arithmetic: the integral is well formed and evaluates cleanly to a wrong answer. The only defence is the check above, run before the setup rather than after it.