, the area
enclosed by a circle
is . Here the Greek letter
represents the constant
ratio of the circumference
of any circle to its diameter
, approximately equal to 3.1416.
One method of deriving this formula, which originated with Archimedes
, involves viewing the circle as the limit
of a sequence of regular polygon
s. The area of a regular polygon is half its perimeter
multiplied by the distance from its center to its sides
, and the corresponding formula–that the area is half the perimeter times the radius–namely, , holds in the limit for a circle.
Although often referred to as the area of a circle in informal contexts, strictly speaking the term ''disk
'' refers to the interior of the circle, while ''circle'' is reserved for the boundary only, which is a curve
and covers no area itself. Therefore, the area of a disk is the more precise phrase for the area enclosed by a circle.
Modern mathematics can obtain the area using the methods of integral calculus
or its more sophisticated offspring, real analysis
. However, the area of a disk was studied by the Ancient Greeks
. Eudoxus of Cnidus
in the fifth century B.C. had found that the area of a disk is proportional to its radius squared. Archimedes
used the tools of Euclidean geometry
to show that the area inside a circle is equal to that of a right triangle
whose base has the length of the circle's circumference and whose height equals the circle's radius in his book ''Measurement of a Circle
''. The circumference is 2''r'', and the area of a triangle is half the base times the height, yielding the area ''r''2
for the disk. Prior to Archimedes, Hippocrates of Chios
was the first to show that the area of a disk is proportional to the square of its diameter, as part of his quadrature of the lune of Hippocrates
but did not identify the constant of proportionality
A variety of arguments have been advanced historically to establish the equation
to varying degrees of mathematical rigor. The most famous of these is Archimedes' method of exhaustion
, one of the earliest uses of the mathematical concept of a limit
, as well as the origin of Archimedes' axiom
which remains part of the standard analytical treatment of the real number system
. The original proof of Archimedes is not rigorous by modern standards, because it assumes that we can compare the length of arc of a circle to the length of a secant and a tangent line, and similar statements about the area, as geometrically evident.
The area of a regular polygon
is half its perimeter times the apothem
. As the number of sides of the regular polygon increases, the polygon tends to a circle, and the apothem tends to the radius. This suggests that the area of a disk is half the circumference of its bounding circle times the radius.
Following Archimedes' argument in ''The Measurement of a Circle'' (c. 260 BCE), compare the area enclosed by a circle to a right triangle whose base has the length of the circle's circumference and whose height equals the circle's radius. If the area of the circle is not equal to that of the triangle, then it must be either greater or less. We eliminate each of these by contradiction, leaving equality as the only possibility. We use regular polygon
s in the same way.
Suppose that the area ''C'' enclosed by the circle is greater than the area ''T'' = 1
''cr'' of the triangle. Let ''E'' denote the excess amount. Inscribe
a square in the circle, so that its four corners lie on the circle. Between the square and the circle are four segments. If the total area of those gaps, ''G''4
, is greater than ''E'', split each arc in half. This makes the inscribed square into an inscribed octagon, and produces eight segments with a smaller total gap, ''G''8
. Continue splitting until the total gap area, ''Gn
'', is less than ''E''. Now the area of the inscribed polygon, ''Pn
'' = ''C'' − ''Gn
'', must be greater than that of the triangle.
But this forces a contradiction, as follows. Draw a perpendicular from the center to the midpoint of a side of the polygon; its length, ''h'', is less than the circle radius. Also, let each side of the polygon have length ''s''; then the sum of the sides, ''ns'', is less than the circle circumference. The polygon area consists of ''n'' equal triangles with height ''h'' and base ''s'', thus equals 1
''nhs''. But since ''h'' < ''r'' and ''ns'' < ''c'', the polygon area must be less than the triangle area, 1
''cr'', a contradiction. Therefore, our supposition that ''C'' might be greater than ''T'' must be wrong.
Suppose that the area enclosed by the circle is less than the area ''T'' of the triangle. Let ''D'' denote the deficit amount. Circumscribe a square, so that the midpoint of each edge lies on the circle. If the total area gap between the square and the circle, ''G''4
, is greater than ''D'', slice off the corners with circle tangents to make a circumscribed octagon, and continue slicing until the gap area is less than ''D''. The area of the polygon, ''Pn
'', must be less than ''T''.
This, too, forces a contradiction. For, a perpendicular to the midpoint of each polygon side is a radius, of length ''r''. And since the total side length is greater than the circumference, the polygon consists of ''n'' identical triangles with total area greater than ''T''. Again we have a contradiction, so our supposition that ''C'' might be less than ''T'' must be wrong as well.
Therefore, it must be the case that the area enclosed by the circle is precisely the same as the area of the triangle. This concludes the proof.
Following Satō Moshun and Leonardo da Vinci
, we can use inscribed regular polygons in a different way. Suppose we inscribe a hexagon
. Cut the hexagon into six triangles by splitting it from the center. Two opposite triangles both touch two common diameters; slide them along one so the radial edges are adjacent. They now form a parallelogram
, with the hexagon sides making two opposite edges, one of which is the base, ''s''. Two radial edges form slanted sides, and the height, ''h'' is equal to its apothem
(as in the Archimedes proof). In fact, we can also assemble all the triangles into one big parallelogram by putting successive pairs next to each other. The same is true if we increase it to eight sides and so on. For a polygon with 2''n'' sides, the parallelogram will have a base of length ''ns'', and a height ''h''. As the number of sides increases, the length of the parallelogram base approaches half the circle circumference, and its height approaches the circle radius. In the limit, the parallelogram becomes a rectangle with width ''r'' and height ''r''.
There are various equivalent definitions of the constant π. The conventional definition in pre-calculus geometry is the ratio of the circumference of a circle to its diameter:
However, because the circumference of a circle is not a primitive analytical concept, this definition is not suitable in modern rigorous treatments. A standard modern definition is that is equal to twice the least positive root of the cosine
function or, equivalently, the half-period of the sine
(or cosine) function. The cosine function can be defined either as a power series
, or as the solution of a certain differential equation
. This avoids any reference to circles in the definition of , so that statements about the relation of to the circumference and area of circles are actually theorems, rather than definitions, that follow from the analytical definitions of concepts like "area" and "circumference".
The analytical definitions are seen to be equivalent, if it is agreed that the circumference of the circle is measured as a rectifiable curve
by means of the integral
The integral appearing on the right is an abelian integral
whose value is a half-period of the sine
function, equal to . Thus
is seen to be true as a theorem.
Several of the arguments that follow use only concepts from elementary calculus to reproduce the formula
, but in many cases to regard these as actual proofs, they rely implicitly on the fact that one can develop trigonometric functions and the fundamental constant in a way that is totally independent of their relation to geometry. We have indicated where appropriate how each of these proofs can be made totally independent of all trigonometry, but in some cases that requires more sophisticated mathematical ideas than those afforded by elementary calculus.
Using calculus, we can sum the area incrementally, partitioning the disk into thin concentric rings like the layers of an onion
. This is the method of shell integration
in two dimensions. For an infinitesimally thin ring of the "onion" of radius ''t'', the accumulated area is 2''t dt'', the circumferential length of the ring times its infinitesimal width (one can approximate this ring by a rectangle with width=2''t'' and height=''dt''). This gives an elementary integral for a disk of radius ''r''.
It is rigorously justified by the multivariate substitution rule
in polar coordinates. Namely, the area is given by a double integral
of the constant function 1 over the disk itself. If ''D'' denotes the disk, then the double integral can be computed in polar coordinates
which is the same result as obtained above.
An equivalent rigorous justification, without relying on the special coordinates of trigonometry, uses the coarea formula
. Define a function
. Note ρ is a Lipschitz function
is a unit vector
). Let ''D'' be the disc
. We will show that
is the two-dimensional Lebesgue measure in
. We shall assume that the one-dimensional Hausdorff measure
of the circle
, the circumference of the circle of radius ''r''. (This can be taken as the definition of circumference.) Then, by the coarea formula,
Similar to the onion proof outlined above, we could exploit calculus in a different way in order to arrive at the formula for the area of a disk. Consider unwrapping the concentric circles to straight strips. This will form a right angled triangle with r as its height and 2r (being the outer slice of onion) as its base.
Finding the area of this triangle will give the area of the disk
The opposite and adjacent angles for this triangle are respectively in degrees 9.0430611..., 80.956939... and in radians 0.1578311... , 1.4129651....
Explicitly, we imagine dividing up a circle into triangles, each with a height equal to the circle's radius and a base that is infinitesimally small. The area of each of these triangles is equal to
. By summing up (integrating) all of the areas of these triangles, we arrive at the formula for the circle's area:
It too can be justified by a double integral of the constant function 1 over the disk by reversing the order of integration
and using a change of variables in the above iterated integral:
Making the substitution
converts the integral to
which is the same as the above result.
The triangle proof can be reformulated as an application of Green's theorem
in flux-divergence form (i.e. a two-dimensional version of the divergence theorem
), in a way that avoids all mention of trigonometry and the constant . Consider the vector field
in the plane. So the divergence
of r is equal to two, and hence the area of a disc ''D'' is equal to
By Green's theorem, this is the same as the outward flux of r across the circle bounding ''D'':
where n is the unit normal and ''ds'' is the arc length measure. For a circle of radius ''R'' centered at the origin, we have
, so the above equality is
The integral of ''ds'' over the whole circle
is just the arc length, which is its circumference, so this shows that the area ''A'' enclosed by the circle is equal to
times the circumference of the circle.
Another proof that uses triangles considers the area enclosed by a circle to be made up of an infinite number of triangles (i.e. the triangles each have an angle of ' at the centre of the circle), each with an area of ' (derived from the expression for the area of a triangle: '). Note that ''≈ '' due to small angle approximation
. Through summing the areas of the triangles, the expression for the area of the circle can therefore be found:
Note that the area of a semicircle of radius r can be computed by the integral
frame|A semicircle of radius ''r''
By trigonometric substitution
, we substitute
The last step follows since the trigonometric identity
have equal integrals over the interval