What is the area of one slice? AB & BC

Slice the solid perpendicular to the $x$-axis. Each slice is a flat plate of area $A(x)$ and thickness $dx$, so its volume is $A(x)\,dx$ and

$$V = \int_a^b A(x)\,dx.$$

That is the entire theory. What changes between problems is $A(x)$, and $A(x)$ is determined by two things: the shape of the cross section, which the problem states, and the length $s$ of the slice's base, which comes from the region.

Check the units and the structure becomes obvious. $A$ is an area, $dx$ is a length, and their product is a volume. The same reasoning slices perpendicular to the $y$-axis instead, giving $V = \int_c^d A(y)\,dy$, and the choice is made the same way as in 8.5: slice in the direction where the boundaries keep their identity.

The slice stands on the base region, so its footprint is the segment cut out of the region at position $x$. The length of that segment is

$$s(x) = y_{\text{top}} - y_{\text{bottom}},$$

which is 8.4's integrand, reused. If the region runs between $y = x$ and $y = x^{2}$ on $[0, 1]$, then $s = x - x^{2}$, and at $x = \frac{1}{2}$ that is $\frac{1}{4}$.

The error CA10 names is taking $s$ from a single curve, writing $s = x$ here. That is correct only when the lower boundary is the $x$-axis, and it is worth seeing why the mistake is so easy: most introductory examples do have the axis as the lower boundary, so $s = f(x)$ works until it suddenly does not.

1. Name both boundaries before writing $s$. If one of them is $y = 0$, say so out loud rather than assuming it.
2. Check $s$ at one interior value. At $x = \frac{1}{2}$ the region runs from $\frac{1}{4}$ to $\frac{1}{2}$, so $s = \frac{1}{4}$. Any expression not giving $\frac{1}{4}$ there is describing a different segment.
3. Check $s$ at the ends. Where the boundaries meet, $s$ should be zero. Here $s = 0$ at $x = 0$ and $x = 1$, which is what makes the solid come to a point at both ends.

For a square cross section, $A = s^{2}$. For the running example,

$$V = \int_0^1 (x - x^{2})^{2}\,dx = \int_0^1 \big(x^{2} - 2x^{3} + x^{4}\big)\,dx = \frac{1}{3} - \frac{1}{2} + \frac{1}{5} = \frac{1}{30}.$$

Two errors live in that first step, and both are CA9.

1. Not squaring at all. $\int_0^1 (x - x^{2})\,dx = \frac{1}{6}$, which is the area of the base, not the volume of anything.
2. Squaring the terms separately. $\int_0^1 (x^{2} - x^{4})\,dx = \frac{2}{15}$, which uses $(a - b)^{2} = a^{2} - b^{2}$. That identity is false, and it will reappear in 8.11 as the washer error, where it costs the cross term again.

Rectangles are squares with one more given. If the problem says the height is $k$ times the base, then $A = s \cdot ks = k s^{2}$, and the constant rides through the integral untouched. Cross sections that are rectangles "of height $3$" are different again: there $A = 3s$, with no squaring, because only one dimension comes from the region.

Read the sentence carefully enough to know which of those two you have. "Height equal to three times the base" is $3s^{2}$; "height $3$" is $3s$.

Every remaining topic in this unit is the same integral with a different $A$:

1. 8.8: triangles and semicircles, so $A$ is $\frac{1}{2}s \cdot h$ or $\frac{\pi}{8}s^{2}$.
2. 8.9 and 8.10: discs, so $A = \pi r^{2}$ and the slice is a circle rather than a square.
3. 8.11 and 8.12: washers, so $A = \pi(R^{2} - r^{2})$ and the slice is a circle with a hole.

Seeing that early is worth more than any individual formula, because it converts six topics into one method with six values of $A$. The recurring work is always the same: identify the two boundaries, form the distance between them, and only then apply the shape's area formula.

One consequence is worth stating now. A solid of revolution is not a separate idea from a solid with known cross sections; it is the case where the cross sections happen to be circles, and the radius happens to be a distance measured from the axis of revolution. Topics 8.9 through 8.12 are this topic with $A = \pi r^{2}$.