Area of a circle

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.[3]

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 polygons in the same way.

Circle with square and octagon inscribed, showing area gap

Suppose that the area C enclosed by the circle is greater than the area T = ¹⁄₂cr 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, G₄, 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, G₈. Continue splitting until the total gap area, G_n, is less than E. Now the area of the inscribed polygon, P_n = C − G_n, must be greater than that of the triangle.

${\displaystyle {\begin{aligned}E&{}=C-T\\&{}>G_{n}\\P_{n}&{}=C-G_{n}\\&{}>C-E\\P_{n}&{}>T\end{aligned}}}$

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 ¹⁄₂nhs. But since h < r and ns < c, the polygon area must be less than the triangle area, ¹⁄₂cr, a contradiction. Therefore, our supposition that C might be greater than T must be wrong.

Circle with square and octagon circumscribed, showing area gap

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, G₄, 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, P_n, must be less than T.

${\displaystyle {\begin{aligned}D&{}=T-C\\&{}>G_{n}\\P_{n}&{}=C+G_{n}\\&{}<C+D\\P_{n}&{}<T\end{aligned}}}$

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.

Circle area by rearrangement

Graphs of side, s; apothem, a; and area, A of regular polygons of n sides and circumradius 1, with the base, b of a rectangle with the same area. The green line shows the case n = 6.

Following Satō Moshun (Smith & Mikami 1914, pp. 130–132), Nicholas of Cusa[4] and Leonardo da Vinci (Beckmann 1976, p. 19), 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 2n 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 πr and height r.

Unit disk area by rearranging n polygons.

polygon			parallelogram
n	side		base	height	area
4	1.4142136		2.8284271	0.7071068	2.0000000
6	1.0000000		3.0000000	0.8660254	2.5980762
8	0.7653669		3.0614675	0.9238795	2.8284271
10	0.6180340		3.0901699	0.9510565	2.9389263
12	0.5176381		3.1058285	0.9659258	3.0000000
14	0.4450419		3.1152931	0.9749279	3.0371862
16	0.3901806		3.1214452	0.9807853	3.0614675
96	0.0654382		3.1410320	0.9994646	3.1393502
∞	1/∞		π	1	π

Area of the disk via ring integration

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 2πt dt, the circumferential length of the ring times its infinitesimal width (one can approximate this ring by a rectangle with width=2πt and height=dt). This gives an elementary integral for a disk of radius r.

${\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\int _{0}^{r}2\pi t\,dt\\&{}=2\pi \left[{\frac {t^{2}}{2}}\right]_{0}^{r}\\&{}=\pi r^{2}.\end{aligned}}}$

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 as follows:

${\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\iint _{D}1\ d(x,y)\\&{}=\iint _{D}t\ dt\ d\theta \\&{}=\int _{0}^{r}\int _{0}^{2\pi }t\ d\theta \ dt\\&{}=\int _{0}^{r}\left[t\theta \right]_{0}^{2\pi }dt\\&{}=\int _{0}^{r}2\pi t\,dt\\\end{aligned}}}$

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 ${\displaystyle \rho$ :\mathbb {R} ^{2}\to \mathbb {R} } by ${\textstyle \rho (x,y)={\sqrt {x^{2}+y^{2}}}}$. Note ρ is a Lipschitz function whose gradient is a unit vector ${\displaystyle |\nabla \rho |=1}$ (almost everywhere). Let D be the disc ${\displaystyle \rho <1}$ in ${\displaystyle \mathbb {R} ^{2}}$. We will show that ${\displaystyle {\mathcal {L}}^{2}(D)=\pi }$, where ${\displaystyle {\mathcal {L}}^{2}}$ is the two-dimensional Lebesgue measure in ${\displaystyle \mathbb {R} ^{2}}$. We shall assume that the one-dimensional Hausdorff measure of the circle ${\displaystyle \rho =r}$ is ${\displaystyle 2\pi r}$, the circumference of the circle of radius r. (This can be taken as the definition of circumference.) Then, by the coarea formula,

${\displaystyle {\begin{aligned}{\mathcal {L}}^{2}(D)&=\iint _{D}|\nabla \rho |\,d{\mathcal {L}}^{2}\\&=\int _{\mathbb {R} }{\mathcal {H}}^{1}(\rho ^{-1}(r)\cap D)\,dr\\&=\int _{0}^{1}{\mathcal {H}}^{1}(\rho ^{-1}(r))\,dr\\&=\int _{0}^{1}2\pi r\,dr=\pi .\end{aligned}}}$

Circle unwrapped to form a triangle

The circle and the triangle are equal in area.

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 2πr (being the outer slice of onion) as its base.

Finding the area of this triangle will give the area of the disk

${\displaystyle {\begin{aligned}{\text{Area}}&{}={\frac {1}{2}}\cdot {\text{base}}\cdot {\text{height}}\\&{}={\frac {1}{2}}\cdot 2\pi r\cdot r\\&{}=\pi r^{2}\end{aligned}}}$

The opposite and adjacent angles for this triangle are respectively in degrees 9.0430611..., 80.956939... and in radians 0.1578311... OEIS: A233527, 1.4129651...OEIS: A233528.

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 ${\displaystyle 1/2\cdot r\cdot du}$. By summing up (integrating) all of the areas of these triangles, we arrive at the formula for the circle's area:

${\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\int _{0}^{2\pi r}{\frac {1}{2}}r\,du\\&{}=\left[{\frac {1}{2}}ru\right]_{0}^{2\pi r}\\&{}=\pi r^{2}.\end{aligned}}}$

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:

${\displaystyle {\begin{aligned}\mathrm {Area} (r)&{}=\iint _{D}1\ d(x,y)\\&{}=\iint _{D}t\ dt\ d\theta \\&{}=\int _{0}^{2\pi }\int _{0}^{r}t\ dt\ d\theta \\&{}=\int _{0}^{2\pi }{\frac {1}{2}}r^{2}\ d\theta \\\end{aligned}}}$

Making the substitution ${\displaystyle u=r\theta ,\ du=r\ d\theta }$ converts the integral to

${\displaystyle \int _{0}^{2\pi r}{\frac {1}{2}}{\frac {r^{2}}{r}}du=\int _{0}^{2\pi r}{\frac {1}{2}}r\ du}$

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 π. Consider the vector field ${\displaystyle \mathbf {r} =x\mathbf {i} +y\mathbf {j} }$ in the plane. So the divergence of r is equal to two, and hence the area of a disc D is equal to

${\displaystyle A={\frac {1}{2}}\iint _{D}\operatorname {div} \mathbf {r} \,dA.}$

By Green's theorem, this is the same as the outward flux of r across the circle bounding D:

${\displaystyle A={\frac {1}{2}}\oint _{\partial D}\mathbf {r} \cdot \mathbf {n} \,ds}$

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 ${\displaystyle |\mathbf {r} |=R}$ and ${\displaystyle \mathbf {n} =\mathbf {r} /R}$, so the above equality is

${\displaystyle A={\frac {1}{2}}\oint _{\partial D}\mathbf {r} \cdot {\frac {\mathbf {r} }{R}}\,ds={\frac {R}{2}}\oint _{\partial D}\,ds.}$

The integral of ds over the whole circle ${\displaystyle \partial D}$ is just the arc length, which is its circumference, so this shows that the area A enclosed by the circle is equal to ${\displaystyle R/2}$ 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 d𝜃 at the centre of the circle), each with an area of 1/2 · r² · d𝜃 (derived from the expression for the area of a triangle: 1/2 · a · b · sin𝜃 = 1/2 · r · r · sin(d𝜃) = 1/2 · r² · d𝜃). Note that sin(d𝜃) ≈ d𝜃 due to small angle approximation. Through summing the areas of the triangles, the expression for the area of the circle can therefore be found:

$${\displaystyle {\begin{aligned}\mathrm {Area} &{}=\int _{0}^{2\pi }{\frac {1}{2}}r^{2}\,d\theta \\&{}=\left[{\frac {1}{2}}r^{2}\theta \right]_{0}^{2\pi }\\&{}=\pi r^{2}.\end{aligned}}}$$

Note that the area of a semicircle of radius r can be computed by the integral ${\textstyle \int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\,dx}$.

A semicircle of radius r

By trigonometric substitution, we substitute ${\displaystyle x=r\sin \theta }$, hence ${\displaystyle dx=r\cos \theta \,d\theta .}$

$${\displaystyle {\begin{aligned}\int _{-r}^{r}{\sqrt {r^{2}-x^{2}}}\,dx&=\int _{-{\frac {\pi }{2}}}^{\frac {\pi }{2}}{\sqrt {r^{2}\left(1-\sin ^{2}\theta \right)}}\cdot r\cos \theta \,d\theta \\&=2r^{2}\int _{0}^{\frac {\pi }{2}}\cos ^{2}\theta \,d\theta \\&={\frac {\pi r^{2}}{2}}.\end{aligned}}}$$

The last step follows since the trigonometric identity ${\displaystyle \cos(\theta )=\sin(\pi /2-\theta )}$ implies that ${\displaystyle \cos ^{2}\theta }$ and ${\displaystyle \sin ^{2}\theta }$ have equal integrals over the interval ${\displaystyle [0,\pi /2]}$, using integration by substitution. But on the other hand, since ${\displaystyle \cos ^{2}\theta +\sin ^{2}\theta =1}$, the sum of the two integrals is the length of that interval, which is ${\displaystyle \pi /2}$. Consequently, the integral of ${\displaystyle \cos ^{2}\theta }$ is equal to half the length of that interval, which is ${\displaystyle \pi /4}$.

Therefore, the area of a circle of radius r, which is twice the area of the semi-circle, is equal to ${\displaystyle 2\cdot {\frac {\pi r^{2}}{2}}=\pi r^{2}}$.

This particular proof may appear to beg the question, if the sine and cosine functions involved in the trigonometric substitution are regarded as being defined in relation to circles. However, as noted earlier, it is possible to define sine, cosine, and π in a way that is totally independent of trigonometry, in which case the proof is valid by the change of variables formula and Fubini's theorem, assuming the basic properties of sine and cosine (which can also be proved without assuming anything about their relation to circles).

Given a circle, let u_n be the perimeter of an inscribed regular n-gon, and let U_n be the perimeter of a circumscribed regular n-gon. Then u_n and U_n are lower and upper bounds for the circumference of the circle that become sharper and sharper as n increases, and their average (u_n + U_n)/2 is an especially good approximation to the circumference. To compute u_n and U_n for large n, Archimedes derived the following doubling formulae:

${\displaystyle u_{2n}={\sqrt {U_{2n}u_{n}}}}$   (geometric mean), and

${\displaystyle U_{2n}={\frac {2U_{n}u_{n}}{U_{n}+u_{n}}}}$    (harmonic mean).

Starting from a hexagon, Archimedes doubled n four times to get a 96-gon, which gave him a good approximation to the circumference of the circle.

In modern notation, we can reproduce his computation (and go further) as follows. For a unit circle, an inscribed hexagon has u₆ = 6, and a circumscribed hexagon has U₆ = 4√3. Doubling seven times yields

Archimedes doubling seven times; n = 6×2^k.

k	n	un	Un	un + Un/4
0	6	6.0000000	6.9282032	3.2320508
1	12	6.2116571	6.4307806	3.1606094
2	24	6.2652572	6.3193199	3.1461443
3	48	6.2787004	6.2921724	3.1427182
4	96	6.2820639	6.2854292	3.1418733
5	192	6.2829049	6.2837461	3.1416628
6	384	6.2831152	6.2833255	3.1416102
7	768	6.2831678	6.2832204	3.1415970

(Here u_n + U_n/2 approximates the circumference of the unit circle, which is 2π, so u_n + U_n/4 approximates π.)

The last entry of the table has ³⁵⁵⁄₁₁₃ as one of its best rational approximations; i.e., there is no better approximation among rational numbers with denominator up to 113. The number ³⁵⁵⁄₁₁₃ is also an excellent approximation to π, attributed to Chinese mathematician Zu Chongzhi, who named it Milü.[5] This approximation is better than any other rational number with denominator less than 16,604.[6]

Snell proposed (and Huygens proved) a tighter bound than Archimedes':

${\displaystyle n{\frac {3\sin {\frac {\pi }{n}}}{2+\cos {\frac {\pi }{n}}}}<\pi <n\left(2\sin {\frac {\pi }{3n}}+\tan {\frac {\pi }{3n}}\right).}$

This for n = 48 gives a better approximation (about 3.14159292) than Archimedes' method for n = 768.

Circle with similar triangles: circumscribed side, inscribed side and complement, inscribed split side and complement

Let one side of an inscribed regular n-gon have length s_n and touch the circle at points A and B. Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem, this is a right triangle with right angle at B. Let the length of A′B be c_n, which we call the complement of s_n; thus c_n²+s_n² = (2r)². Let C bisect the arc from A to B, and let C′ be the point opposite C on the circle. Thus the length of CA is s_2n, the length of C′A is c_2n, and C′CA is itself a right triangle on diameter C′C. Because C bisects the arc from A to B, C′C perpendicularly bisects the chord from A to B, say at P. Triangle C′AP is thus a right triangle, and is similar to C′CA since they share the angle at C′. Thus all three corresponding sides are in the same proportion; in particular, we have C′A : C′C = C′P : C′A and AP : C′A = CA : C′C. The center of the circle, O, bisects A′A, so we also have triangle OAP similar to A′AB, with OP half the length of A′B. In terms of side lengths, this gives us

${\displaystyle {\begin{aligned}c_{2n}^{2}&{}=\left(r+{\frac {1}{2}}c_{n}\right)2r\\c_{2n}&{}={\frac {s_{n}}{s_{2n}}}.\end{aligned}}}$

In the first equation C′P is C′O+OP, length r+¹⁄₂c_n, and C′C is the diameter, 2r. For a unit circle we have the famous doubling equation of Ludolph van Ceulen,

${\displaystyle c_{2n}={\sqrt {2+c_{n}}}.}$

If we now circumscribe a regular n-gon, with side A″B″ parallel to AB, then OAB and OA″B″ are similar triangles, with A″B″ : AB = OC : OP. Call the circumscribed side S_n; then this is S_n : s_n = 1 : ¹⁄₂c_n. (We have again used that OP is half the length of A′B.) Thus we obtain

${\displaystyle c_{n}=2{\frac {s_{n}}{S_{n}}}.}$

Call the inscribed perimeter u_n = ns_n, and the circumscribed perimeter U_n = nS_n. Then combining equations, we have

${\displaystyle c_{2n}={\frac {s_{n}}{s_{2n}}}=2{\frac {s_{2n}}{S_{2n}}},}$

so that

${\displaystyle u_{2n}^{2}=u_{n}U_{2n}.}$

This gives a geometric mean equation.

We can also deduce

${\displaystyle 2{\frac {s_{2n}}{S_{2n}}}{\frac {s_{n}}{s_{2n}}}=2+2{\frac {s_{n}}{S_{n}}},}$

or

${\displaystyle {\frac {2}{U_{2n}}}={\frac {1}{u_{n}}}+{\frac {1}{U_{n}}}.}$

This gives a harmonic mean equation.