Frustum

The volume formula of a frustum of a square pyramid was introduced by the ancient Egyptian mathematics in what is called the Moscow Mathematical Papyrus, written in the 13th dynasty (c. 1850 BC):

${\displaystyle V={\tfrac {1}{3}}h\left(a^{2}+ab+b^{2}\right).}$

where a and b are the base and top side lengths of the truncated pyramid, and h is the height. The Egyptians knew the correct formula for obtaining the volume of a truncated square pyramid, but no proof of this equation is given in the Moscow papyrus.

The volume of a conical or pyramidal frustum is the volume of the solid before slicing the apex off, minus the volume of the apex:

${\displaystyle V={\frac {h_{1}B_{1}-h_{2}B_{2}}{3}}}$

where B₁ is the area of one base, B₂ is the area of the other base, and h₁, h₂ are the perpendicular heights from the apex to the planes of the two bases.

Considering that

${\displaystyle {\frac {B_{1}}{h_{1}^{2}}}={\frac {B_{2}}{h_{2}^{2}}}={\frac {\sqrt {B_{1}B_{2}}}{h_{1}h_{2}}}=\alpha }$,

the formula for the volume can be expressed as a product of this proportionality α/3 and a difference of cubes of heights h₁ and h₂ only.

${\displaystyle V={\frac {h_{1}\alpha h_{1}^{2}-h_{2}\alpha h_{2}^{2}}{3}}={\frac {\alpha }{3}}(h_{1}^{3}-h_{2}^{3})}$

By factoring the difference of two cubes, a³ − b³ = (a − b)(a² + ab + b²), one gets h₁ − h₂ = h, the height of the frustum, and αh₁² + h₁h₂ + h₂²/3.

Distributing α and substituting from its definition, the Heronian mean of areas B₁ and B₂ is obtained. The alternative formula is therefore

${\displaystyle V={\frac {h}{3}}\left(B_{1}+{\sqrt {B_{1}B_{2}}}+B_{2}\right)}$.

Heron of Alexandria is noted for deriving this formula and with it encountering the imaginary unit, the square root of negative one.[2]

In particular, the volume of a circular cone frustum is

${\displaystyle V={\frac {\pi h}{3}}\left(r_{1}^{2}+r_{1}r_{2}+r_{2}^{2}\right)}$

where r₁, r₂ are the radii of the two bases.

Pyramidal frustum

The volume of a pyramidal frustum whose bases are n-sided regular polygons is

${\displaystyle V={\frac {nh}{12}}\left(a_{1}^{2}+a_{1}a_{2}+a_{2}^{2}\right)\cot {\frac {\pi }{n}}}$

where a₁ and a₂ are the sides of the two bases.

Conical frustum

3D model of a conical frustum.

For a right circular conical frustum[3][4]

${\displaystyle {\begin{aligned}{\text{Lateral surface area}}&=\pi \left(r_{1}+r_{2}\right)s\\&=\pi \left(r_{1}+r_{2}\right){\sqrt {\left(r_{1}-r_{2}\right)^{2}+h^{2}}}\end{aligned}}}$

and

${\displaystyle {\begin{aligned}{\text{Total surface area}}&=\pi \left(\left(r_{1}+r_{2}\right)s+r_{1}^{2}+r_{2}^{2}\right)\\&=\pi \left(\left(r_{1}+r_{2}\right){\sqrt {\left(r_{1}-r_{2}\right)^{2}+h^{2}}}+r_{1}^{2}+r_{2}^{2}\right)\end{aligned}}}$

where r₁ and r₂ are the base and top radii respectively, and s is the slant height of the frustum.

The surface area of a right frustum whose bases are similar regular n-sided polygons is

${\displaystyle A={\frac {n}{4}}\left[\left(a_{1}^{2}+a_{2}^{2}\right)\cot {\frac {\pi }{n}}+{\sqrt {\left(a_{1}^{2}-a_{2}^{2}\right)^{2}\sec ^{2}{\frac {\pi }{n}}+4h^{2}\left(a_{1}+a_{2}\right)^{2}}}\right]}$

where a₁ and a₂ are the sides of the two bases.