AP Calculus AB and BC

Known Cross-Sections vs Solids of Revolution

Both methods integrate cross-sectional area along an axis. A solid of revolution always has circular or ring-shaped slices because it is swept by rotation. A solid with known cross-sections can have squares, semicircles, or triangles instead, and the problem tells you which.

Solid of revolution

Use when: The problem says a region is rotated or revolved about a line.

Known cross-sections

Use when: The problem describes slices of a stated shape built on a base region.

Side by side

	Revolution	Known cross-sections
Slice shape	Circle or ring	Whatever the problem states
Formula	$V = \pi\int R^2\,dx$	$V = \int A(x)\,dx$
Key quantity	The radius	The base segment width
Involves $\pi$	Always	Only for circular shapes

Revolution is the special case where $A(x)$ happens to be $\pi R^2$ . Once you see that, both problems reduce to the same job: write the area of a representative slice as a function of position, then integrate.

For known cross-sections the base segment width does the work. Squares give $w^2$ , equilateral triangles give $\frac{\sqrt{3}}{4}w^2$ , and semicircles give $\frac{\pi w^2}{8}$ because the segment is the diameter.

The semicircle trap

When cross-sections are semicircles the base segment is the diameter, so the radius is half of it. Using the full width as the radius quadruples the answer.

Frequently asked questions

Do I ever need pi for known cross-sections?

Only when the stated shape is circular, such as a semicircle. Squares and triangles produce no pi at all.

In the CED: Unit 8: Applications of Integration