Prism, cylinder, box, and ball

Two measures come with a solid: the volume $V$ of the region it fills, and the surface area $O$ of its boundary. (The letter $O$ is the table's name for the surface; $M$ is reserved for a lateral area when the table splits one off.)

The table lists a cube, a rectangular parallelepiped, a ball, a prism, a cylinder, and a solid annulus. Six names look like six topics. Five of them are one rule — base area times height — with extra freezes. The ball is not that rule. Turn the figures; these are live 3D constructions.

This math handbook writes lengths in the same units as the plane figures in the previous section.

Sweep a base through a height

Two parallel congruent bases, joined by parallelogram sides. Let $G$ be the area of a base and $h$ the corresponding height:

$V=Gh.$

That is the common relation. The table leaves the surface blank; this page does not invent one. The board freezes an extra condition: a regular hexagonal right prism, matching the table sketch. Drag $A$ to change the base; drag $K$ for $h$. Drop two parallel congruent bases and $V=Gh$ is gone.

Freeze a circular base

Keep $V=Gh$ and freeze the base as a disk of radius $r$. Then $G=\pi r^2$, so a right circular cylinder of height $h$ has

$V=\pi r^{2}h,O=M+2\pi r^2,M=2\pi rh.$

The extra freeze is the circle: the wall unrolls to a rectangle, so the table can split off a lateral area $M$. Drag $R$ for $r$ and $B$ for $h$.

Punch a coaxial hole

Keep the cylinder and freeze a coaxial cylindrical hole of inner radius $\rho$ and the same height $h$. The volume is the difference of two cylinder volumes:

$V=\pi h(r^2-{\rho }^2).$

The table leaves the surface blank. This solid annulus is not the torus (a doughnut whose tube has radius $\rho$). Keep $P$ closer to $A$ than $R$ is, so the hole stays inside. Drop the hole and you are back at $\pi r^{2}h$.

Freeze three right edges

Go back to $V=Gh$ and freeze a rectangular base at right angles to the height — the table's rectangular parallelepiped (a cuboid). Then $G=ab$ and three edges $a$, $b$, $c$ meet at right angles:

$V=abc,O=2(ab+bc+ca).$

The extra freeze is those right angles, not a new volume rule. Drag $B$ for $a$, $C$ for $b$, and $H$ for $c$.

Then equal edges

Keep the cuboid and freeze $a=b=c$. Then

$V=a^3,O=6a^2.$

That is a cube. Drag $B$. $A$ stays fixed; the other vertices follow so every edge stays $a$. Two freezes stacked — equal edges on a rectangular box — not a sixth independent solid.

A ball is not an extrusion

A ball of radius $r$ (the table's name for the solid sphere) is not $V=Gh$:

$V=\frac{4}{3}\pi r^3,O=4\pi r^2.$

One radius from a fixed centre, no base to sweep. Drag $A$; $C$ is the centre, so $CA$ is $r$.

Pyramids, cones, and frustums continue the same table — solids that taper, so the cross-section is no longer constant. The spherical cap, torus, and ellipsoid close it. Plane figures live under triangles, quadrilaterals, regular n-gons, and circles, sectors, and ellipses.