{{short description|Primitive way of calculating area}}
{{about|the method of finding the area of a shape using limits|the method of proof|Proof by exhaustion}}

The '''method of exhaustion''' ({{Langx|la|methodus exhaustionis}}) is a method of finding the [[area]] of a [[shape]] by [[Inscribed figure|inscribing]] inside it a [[sequence]] of [[polygon]]s (one at a time) whose [[area]]s [[limit (mathematics)|converge]] to the area of the containing [[shape]]. If the sequence is correctly constructed, the difference in area between the ''n''th polygon and the containing shape will become arbitrarily small as ''n'' becomes large. As this difference becomes arbitrarily small, the possible values for the area of the shape are systematically "exhausted" by the lower bound areas successively established by the sequence members.

The method of exhaustion typically required a form of proof by contradiction, known as ''[[reductio ad absurdum]]''. This amounts to finding an area of a region by first comparing it to the area of a second region, which can be "exhausted" so that its area becomes arbitrarily close to the true area. The proof involves assuming that the true area is greater than the second area, proving that assertion false, assuming it is less than the second area, then proving that assertion false, too.

== History ==
[[File:Grégoire de Saint-Vincent (1584-1667).jpg|thumb|150px|right|Gregory of Saint Vincent]]
The idea originated in the late 5th century BC with [[Antiphon (person)|Antiphon]], although it is not entirely clear how well he understood it.<ref>{{Cite web|url=http://www-history.mcs.st-andrews.ac.uk/Biographies/Antiphon.html|title=Antiphon (480 BC-411 BC)|website=www-history.mcs.st-andrews.ac.uk}}</ref> The theory was made rigorous a few decades later by [[Eudoxus of Cnidus]], who used it to calculate areas and volumes. It was later reinvented in [[Chinese mathematics|China]] by [[Liu Hui]] in the 3rd century AD in order to find the area of a circle.<ref>Dun, Liu. 1966. "[https://books.google.com/books?id=jaQH6_8Ju-MC&pg=PA279 A comparison of Archimedes' and Liu Hui's studies of circles]." Pp. 279–87 in ''Chinese Studies in the History and Philosophy of Science and Technology'' 179, edited by D. Fan, and R. S. Cohen. [[Kluwer Academic|Kluwer Academic Publishers]]. {{ISBN|0-7923-3463-9}}. p. 279.</ref> The first use of the term was in 1647 by [[Grégoire de Saint-Vincent|Gregory of Saint Vincent]] in ''Opus geometricum quadraturae circuli et sectionum''.

The method of exhaustion is seen as a precursor to the methods of [[calculus]]. The development of [[analytical geometry]] and rigorous [[integral calculus]] in the 17th-19th centuries subsumed the method of exhaustion so that it is no longer explicitly used to solve problems. An important alternative approach was [[Cavalieri's principle]], also termed the ''[[method of indivisibles]]'' which eventually evolved into the [[infinitesimal]] calculus of [[Gilles de Roberval|Roberval]], [[Evangelista Torricelli|Torricelli]], [[John Wallis|Wallis]], [[Gottfried Wilhelm Leibniz|Leibniz]], and others.

=== Euclid ===
[[Euclid]] used the method of exhaustion to prove the following six propositions in the 12th book of his ''[[Euclid's Elements|Elements]]''.

'''Proposition 2''': The area of circles is proportional to the square of their [[Diameter|diameters]].<ref>{{cite web|url=http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII2.html|title=Euclid's Elements, Book XII, Proposition 2|website=aleph0.clarku.edu}}</ref>

'''Proposition 5''': The volumes of two tetrahedra of the same height are proportional to the areas of their triangular bases.<ref>{{cite web|url=http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII5.html|title=Euclid's Elements, Book XII, Proposition 5|website=aleph0.clarku.edu}}</ref>

'''Proposition 10''': The volume of a cone is a third of the volume of the corresponding cylinder which has the same base and height.<ref>{{cite web|url=http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII10.html|title=Euclid's Elements, Book XII, Proposition 10|website=aleph0.clarku.edu}}</ref>

'''Proposition 11''': The volume of a cone (or cylinder) of the same height is proportional to the area of the base.<ref>{{cite web|url=http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII11.html|title=Euclid's Elements, Book XII, Proposition 11|website=aleph0.clarku.edu}}</ref>

'''Proposition 12:''' The volume of a cone (or cylinder) that is similar to another is proportional to the cube of the ratio of the diameters of the bases.<ref>{{cite web|url=http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII12.html|title=Euclid's Elements, Book XII, Proposition 12|website=aleph0.clarku.edu}}</ref>

'''Proposition 18''': The volume of a sphere is proportional to the cube of its diameter.<ref>{{cite web|url=http://aleph0.clarku.edu/~djoyce/java/elements/bookXII/propXII18.html|title=Euclid's Elements, Book XII, Proposition 18|website=aleph0.clarku.edu}}</ref>

=== Archimedes ===
{{Main|Pi}}
[[File:Archimedes pi.svg|thumb|right|300px|Archimedes used the method of exhaustion to compute the area inside a circle]]

[[Archimedes]] used the method of exhaustion as a way to compute the area inside a circle by filling the [[circle]] with a sequence of [[polygon]]s with an increasing number of [[Edge (geometry)|sides]] and a corresponding increase in area.  The quotients formed by the area of these polygons divided by the square of the circle radius can be made arbitrarily close to π as the number of polygon sides becomes large, proving that the area inside the circle of radius {{mvar|r}} is {{math|''πr''<sup>2</sup>}}, {{mvar|[[Pi|π]]}} being defined as the ratio of the circumference to the diameter ({{math|''C''/''d''}}).

He also provided the bounds {{math|3&nbsp;+&nbsp;<sup>10</sup>/<sub>71</sub>&nbsp;<&nbsp;''π''&nbsp;<&nbsp;3&nbsp;+&nbsp;<sup>10</sup>/<sub>70</sub>}} (giving a range of {{math|<sup>1</sup>/<sub>497</sub>}}) by comparing the perimeters of the circle with the perimeters of the [[Inscribed figure|inscribed]] and [[Circumscribed circle|circumscribed]] 96-sided regular polygons.

Other results he obtained with the method of exhaustion included:<ref>{{cite book
 | last = Smith
 | first = David E.
 | year = 1958
 | title = History of Mathematics
 | url = https://archive.org/details/historyofmathema0002smit
 | url-access = registration
 | publisher = Dover Publications
 | location = New York
 | isbn = 0-486-20430-8
}}</ref>
* The area bounded by the intersection of a line and a parabola is 4/3 that of the triangle having the same base and height (the [[quadrature of the parabola]]).
* The area of an ellipse is proportional to a rectangle having sides equal to its major and minor axes.
* The volume of a sphere is 4 times that of a cone having a base of the same radius and height equal to this radius.
* The volume of a cylinder having a height equal to its diameter is 3/2 that of a sphere having the same diameter.
* The area bounded by one [[Archimedean spiral|spiral]] rotation and a line is 1/3 that of the circle having a radius equal to the line segment length.
* Use of the method of exhaustion also led to the successful evaluation of an [[infinite geometric series]] (for the first time).

=== Others ===
[[Galileo Galilei]] used the method of exhaustion to find the centre of mass of a truncated cone.<ref>{{Cite book |last=Heilbron |first=John |year=2010 |title=Galileo |publisher=Oxford University Press |page=36 |isbn=978-0-19-958352-2}}</ref>

Shortly before the development of modern calculus, [[Christopher Wren]] employed the method of exhaustion to discover the exact arc length of the [[cycloid]].<ref>{{cite journal |last1=Whiteside |first1=Derek T. |title=Wren the Mathematician |journal=[[Notes and Records]] |publisher=Royal Society |location=London |year=1960 |volume=15 |pages=107–111 |doi=10.1098/rsnr.1960.0010}}</ref>

== Example 1: The area of an [[Archimedean spiral]] is a third of the enclosing circle ==
[[File:Archimedean Spiral and Enclosing Circle.png|thumb|The area of one turn of the Archimedean spiral <math display="inline">r=\theta</math> is a third of the area of the circle enclosing it.]]
Archimedes computed the area of one turn of the spiral <math display="inline">S</math> given by <math display="inline">r = \theta</math> and found that <math display="inline">a(S) = \frac{1}{3}\pi r^2</math>, that is, one third of <math display="inline">a(C)</math>, the area of the circle enclosing it.

For a sketch of the proof, suppose we wish to show that <math display="inline">a(S) = \frac{1}{3}a(C)</math>. By way of contradiction, assume that <math display="inline">a(S) < \frac{1}{3}a(C)</math>. Divide the interval <math display="inline">[0, 2\pi]</math> into <math display="inline">n</math> equal pieces <math display="inline">\theta=\frac{2\pi}{n}</math>, and for each subinterval find the smallest and largest circular sectors enclosing the spiral. See the second image for clarification. Let <math display="inline">P</math> be the set of sectors on the interior of the spiral, and <math display="inline">Q</math> the set of sectors on the exterior. Then, <math display="inline">a(P)</math> is an underestimate for the area of the spiral, and <math display="inline">a(Q)</math> an overestimate. Archimedes was able to show that, for <math display="inline">n</math> sufficiently large, <math display="inline">a(Q)-a(P)<\varepsilon</math> for any <math display="inline">0<\varepsilon</math>.

[[File:Enclosed areas archimedean spiral.png|thumb|For <math display="inline">n=8</math>, the image shows P (blue) and Q (red). The dark grey area is <math display="inline">a(P)</math>, and the dark and light grey areas together represent <math display="inline">a(Q)</math>.]]

Now, define <math display="inline">\varepsilon := \frac{1}{3}a(C)-a(S)</math>. Then we have <math display="inline">\frac{1}{3}a(C)-a(S)>a(Q)-a(P)</math> by the assumption, and thus <math display="inline">\frac{1}{3}a(C)>a(Q)+a(S)-a(P)>a(Q)</math> since the spiral encloses <math display="inline">P</math>. But we can explicitly calculate the area of <math display="inline">Q</math>, as it equals the sum of the areas of the <math>n</math> exterior circular segments, each of which has area <math display="inline">\frac{\theta}{2} r_i^2,</math> for <math display="inline">i\in\{1,\dots,n\}</math>. That is,

<math display="block">\begin{align}a(Q)&=\frac{\theta}{2} r_1^{2}+\frac{\theta}{2} r_2^{2}+\cdots+\frac{\theta}{2} r_n^{2}\\[5pt]
&=\frac{\theta}{2}\left(\theta^2+(2\theta)^2+\cdots+(n\theta)^2\right) \\[5pt]
&=\frac{\theta^3}{12}n(n+1)(2n+1)
\end{align}</math>

using the formula for the [[Square pyramidal number|sum of squares]], which Archimedes had also discovered.

Thus returning to the inequality we have 

<math display="block">\frac{\theta^{3}}{12}n\left(n+1\right)\left(2n+1\right) < \frac{1}{3}a(C).</math>

Since the radius of the circle was <math display="inline">2\pi</math>, the area of the circle is <math display="inline">a(C)=4 \pi^{3}</math>. Once we plug it to the above inequality, together with <math display="inline">\theta = \frac{2\pi}{n}</math>, we get:

<math display="block">\frac{8 \pi^3}{12 n^3}n\left(n+1\right)\left(2n+1\right) < \frac 43\pi^3.</math>

which gets further reduced to equivalent:

<math display="block">\left(n+1\right)\left(2n+1\right) < 2 n^2.</math>

However, this is false for all positive <math display="inline">n</math>, as the first term on the left side is greater than <math display="inline">n</math> and the second one is greater than <math display="inline">2n</math>, so their product is greater than <math display="inline">2n^2</math>, thus we have reached a contradiction.

The proof that <math display="inline">a(S) > \frac{1}{3}a(C)</math> instead is entirely tantamount.{{huh?|date=September 2025}} Since the area of the spiral is neither less than nor greater than one third the area of the circle, Archimedes concluded that they were equal.<ref>{{cite book |last1=Edwards |first1=Charles |title=The Historical Development of the Calculus |date=1994 |publisher=Springer |isbn=0387943137}}</ref>

== Example 2: Circles are to one another as the squares on their diameters==

This statement that <math display="inline">\frac{a(C_1)}{a(C_2)}=\left(\frac{r_1}{r_2}\right)^2</math> is attributed to Eudoxus, but his exposition does not survive – it is reproduced in Euclid book XII proposition 2. 
[[File:Inscribed polygons in Eudoxus' circle proof.png|thumb| The circles <math display="inline">C_1, C_2</math> and the inscribed polygons <math display="inline">P_1, P_2</math>. Note that <math display="inline">P_1, P_2</math> have the same number of sides. ]]
For a sketch of the proof, assume by way of contradiction that <math display="inline">\frac{a(C_1)}{a(C_2)}>\left(\frac{r_1}{r_2}\right)^2\iff a(C_1)>\left(\frac{r_1}{r_2}\right)^2a(C_2).</math>
Let <math display="inline">P_1, P_2</math> be <math display="inline">n</math>-sided regular convex polygons inscribing <math display="inline">C_1, C_2</math> respectively.
Define <math display="inline">\epsilon:= a(C_1)-\left(\frac{r_1}{r_2}\right)^2a(C_2)</math>. Then, by Euclid book X proposition 1, we can find <math display="inline">N</math> such that whenever <math display="inline">n > N</math>, <math display="inline">a(C_1)-a(P_1)<\epsilon</math>. Thus, using the definition of <math display="inline">\epsilon</math> we get  <math display="inline">a(C_1)-a(P_1)<a(C_1)-\left(\frac{r_1}{r_2}\right)^2a(C_2), \quad \left(\frac{r_1}{r_2}\right)^2a(C_2)<a(P_1).</math>

But for any two regular convex ''polygons'', not circles, it is trivial to show that 
<math display="inline">\frac{a(P_1)}{a(P_2)}=\left(\frac{r_1}{r_2}\right)^2</math>, provided <math>n</math> is fixed. 
Inserting this into the previous statement gives 

<math display="block">\left(\frac{r_1}{r_2}\right)^2a(C_2)<\left(\frac{r_1}{r_2}\right)^2a(P_2)\implies a(C_2)<a(P_2).</math>

However, this is a contradiction, since <math display="inline">P_2  \subset C_2</math>. The next step is to prove that 
<math display="inline">\frac{a(C_1)}{a(C_2)}>\left(\frac{r_1}{r_2}\right)^2</math> is also false. However, the labelling of <math display="inline">C_1, C_2</math> was entirely arbitrary; by relabelling this case also follows without further proof necessary. Therefore, we have that <math display="inline">\frac{a(C_1)}{a(C_2)}=\left(\frac{r_1}{r_2}\right)^2</math>.<ref>{{cite web |last1=Wigderson |first1=Yuval |title=Eudoxus |type=lecture notes |date=2019 |url=https://n.ethz.ch/~ywigderson/math/static/Eudoxus.pdf}}</ref>

== Analysis ==
Both approximating an integral with a [[Riemann sum]] or with the [[trapezoidal rule]] can be seen as modern versions of the method of exhaustion.<ref>{{Cite thesis |last=DeSouza |first=Chelsea E. |title=The Greek Method of Exhaustion: Leading the Way to Modern Integration |date=2012 |access-date=10 March 2026 |degree=M.S. |publisher=The Ohio State University |url=https://etd.ohiolink.edu/acprod/odb_etd/ws/send_file/send?accession=osu1338326658&disposition=inline}}</ref>

== See also ==
* ''[[The Method of Mechanical Theorems]]''
* ''[[Quadrature of the Parabola]]''
* [[Trapezoidal rule]]
* [[Pythagorean theorem]]

== References ==
{{reflist}}

{{Ancient Greek mathematics}}

[[Category:Volume]]
[[Category:Euclidean geometry]]
[[Category:Integral calculus]]
[[Category:History of calculus]]
[[Category:History of mathematics]]
[[Category:5th century BC in Greece]]
[[Category:Ancient Greek mathematics]]