Pyramid, cone, and frustum volume

The prism and cylinder keep a constant cross-section, so $V=Gh$. These six taper. The factor $\frac{1}{3}$ is what that taper costs against the same base and height. $O$ is the table's name for the full surface when it splits one off; $M$ is the lateral piece. Where the table leaves the surface blank, this page does not invent one. Turn the figures.

This math handbook writes angles in radians unless a degree sign is shown.

One third of the prism

Let $G$ be the area of the base and $h$ the corresponding height:

$V=\frac{1}{3}Gh.$

That is the common relation. The board freezes an extra condition: a right pyramid on a square. Drag $A$ to change the base; drag $K$ for $h$. Drop the point apex and you are back at the prism $Gh$.

Freeze a circular base

Keep $V=\frac{1}{3}Gh$ and freeze the base as a disk of radius $r$. Then $G=\pi r^2$, so a right circular cone of height $h$ has

$V=\frac{1}{3}\pi r^{2}h,O=M+\pi r^2,M=\pi rs.$

The extra freeze is the circle: the wall unrolls to a sector of area $M$, where $s$ is a meridian (the slant from the apex to the rim). Drag $R$ for $r$ and $A$ for $h$.

Cut the apex

Keep the pyramid and freeze a cut parallel to the base, so two similar polygons remain. Let $G$ be the lower base, $g$ the upper base, and $h$ the height of the leftover:

$V=\frac{h}{3}(G+\sqrt{Gg}+g).$

When $g=0$ this is $\frac{1}{3}Gh$ again. The board is a square frustum; keep $P$ closer to the origin than $A$ is, so the top stays inside the bottom. Drag $A$ for the lower base, $P$ for the upper base, and $K$ for $h$.

The round cut

Keep that cut and freeze circular bases of radii $r$ and $\rho$. The meridian $s$ joins a point on the lower rim to the matching point on the upper rim:

$V=\frac{\pi h}{3}(r^2+r\rho +{\rho }^2),O=M+\pi (r^2+{\rho }^2),M=\pi s(r+\rho ).$

When $\rho =0$ this is the cone. Drag $R$ for $r$, $P$ for $\rho$, and $B$ for $h$. Keep $P$ closer to $A$ than $R$ is.

Drop similar bases

Two rectangular bases that need not be similar: lower edges $a$ and $b$, upper edges $c$ and $d$, height $h$. The upper rectangle sits over the centre of the lower one. Unlike a capped pyramid, $a/c$ need not equal $b/d$, so the side edges do not meet in one apex. That extra drop is the obelisk.

$V=\frac{h}{6}(ab+(a+c)(b+d)+cd).$

Drag $B$ for $a$, $C$ for $b$, $P$ for $c$, $Q$ for $d$, and $K$ for $h$. Keep the top inside the base.

A ridge instead of an apex

A convex hull whose two end faces are triangles. At the default size those ends are equilateral and the base is a long thin rectangle $a\times b$, matching the table sketch; drag $P$ to shorten the ridge. Let $a$ and $b$ be the base edges, $c$ the upper edge, and $h$ the height:

$V=\frac{\pi }{6}bh(2a+c).$

The live $V$ is the volume of that hull. The extra freeze is the ridge $c$, not a point. Drag $B$ for $a$, $C$ for $b$, $P$ for $c$, and $K$ for $h$.

The spherical cap, torus, and ellipsoid close the table.