Volume

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

${\displaystyle V={\frac {h}{3}}\left(a^{2}+ab+b^{2}\right),}$

where a and b are the base and top side lengths, and h is the height.

The Egyptians knew the correct formula for the volume of such 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 its "apex" off, minus the volume of this "apex":

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

where B₁ and B₂ are the base and top areas, and h₁ and h₂ are the perpendicular heights from the apex to the base and top planes.

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 the third of the product of this proportionality, ${\displaystyle \alpha }$, and of the difference of the cubes of the heights h₁ and h₂ only:

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

By using the identity a³ − b³ = (a − b)(a² + ab + b²), one gets:

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

where h₁ − h₂ = h is the height of the frustum.

Distributing ${\displaystyle \alpha }$ and substituting from its definition, the Heronian mean of areas B₁ and B₂ is obtained:

${\displaystyle {\frac {B_{1}+{\sqrt {B_{1}B_{2}}}+B_{2}}{3}};}$

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.^[4]

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 and r₂ are the base and top radii.

• The volume of a pyramidal frustum whose bases are regular n-gons 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 base and top side lengths.

Pyramidal frustum

Surface area

Conical frustum

3D model of a conical frustum.

For a right circular conical frustum^[5][6]

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

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

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

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