Editing Cone (section)

=== Volume ===
[[File:visual_proof_cone_volume.svg|thumb|[[Proof without words]] that the volume of a cone is a third of a cylinder of equal diameter and height
{|
|valign="top"|{{nowrap|1.}}||A cone and a cylinder have {{nowrap|radius ''r''}} and {{nowrap|height ''h''.}}
|-
|valign="top"|2.||The volume ratio is maintained when the height is scaled to {{nowrap|1=''h' ''= ''r'' &#8730;{{pi}}.}}
|-
|valign="top"|3.||Decompose it into thin slices.
|-
|valign="top"|4.||Using Cavalieri's principle, reshape each slice into a square of the same area.
|-
|valign="top"|5.||The pyramid is replicated twice.
|-
|valign="top"|6.||Combining them into a cube shows that the volume ratio is 1:3.
|}]]
The [[volume]] <math>V</math> of any conic solid is one third of the product of the area of the base <math>A_B</math> and the height <math>h</math><ref name=":0">{{Cite book|url=https://books.google.com/books?id=EN_KAgAAQBAJ|title=Elementary Geometry for College Students|last1=Alexander|first1=Daniel C.|last2=Koeberlein|first2=Geralyn M.|date=2014-01-01|publisher=Cengage|isbn=9781285965901}}</ref> 

<math display=block>V = \frac{1}{3}A_B h.</math>

In modern mathematics, this formula can easily be computed using calculus — it is, up to scaling, the integral

<math display="block">\int x^2 \, dx = \tfrac{1}{3} x^3</math>

Without using calculus, the formula can be proven by comparing the cone to a pyramid and applying [[Cavalieri's principle]] – specifically, comparing the cone to a (vertically scaled) right square pyramid, which forms one third of a cube. This formula cannot be proven without using such infinitesimal arguments – unlike the 2-dimensional formulae for polyhedral area, though similar to the area of the circle – and hence admitted less rigorous proofs before the advent of calculus, with the ancient Greeks using the [[method of exhaustion]]. This is essentially the content of [[Hilbert's third problem]] – more precisely, not all polyhedral pyramids are ''scissors congruent'' (can be cut apart into finite pieces and rearranged into the other), and thus volume cannot be computed purely by using a decomposition argument.<ref>{{Cite book|url=https://books.google.com/books?id=C5fSBwAAQBAJ|title=Geometry: Euclid and Beyond|last=Hartshorne|first=Robin|author-link=Robin Hartshorne|date=2013-11-11|publisher=Springer Science & Business Media|isbn=9780387226767|at=Chapter 27}}</ref>