Area of a circle

The area of a circle with radius r is π r ² . Here the symbol π ( Greek letter pi ) denotes a constant expressing the ratio of the circumference of a circle to its diameter or the area of a circle to the square of its radius . Since the area of a regular polygon is equal to half its perimeter multiplied by apothem (height), and regular polygons tend to circle with increasing number of sides, the area of a circle is equal to half the length of a circle multiplied by radius (i.e. ¹ ⁄ ₂ × 2π r × r ).

History

Modern mathematicians can obtain the area of a circle using methods of integration or material analysis . However, the area of the circle was studied in ancient Greece . Eudoxus of Cnidus in the fifth century BC discovered that the areas of circles are proportional to the squares of their radii. ^{[1] The} great mathematician Archimedes used the methods of Euclidean geometry to show that the area inside the circle is equal to the area of the right triangle , the base of which has the length of the circle, and the height is equal to the radius of the circle, in his book . The circumference is 2 π r , and the area of the triangle is half the height of the base, which gives π r ² . Before Archimedes, Hippocrates of Chios was the first to show that the area of a circle is proportional to the square of its diameter in his attempts to quadrate hippocratic holes ^[2] However, he did not establish a proportionality constant .

Using Polygons

The area of a regular polygon is equal to half the perimeter multiplied by the apothem (height). As the number of sides increases, the polygon tends to a circle, and the apothem tends to a radius. This gives reason to believe that the area of a circle is equal to the product of half the circumference by a radius. ^[3]

Proof of Archimedes

Following Archimedes, we compare the area of a circle with the area of a rectangular triangle, the base of which is equal to the length of the circle, and the height is equal to the radius. If the area of the circle is not equal to the area of the triangle, it should be less or more. We exclude both options, which leaves only one possibility - the areas are equal. For the proof we will use regular polygons .

No more

Circle with inscribed square and octagon. Clearance shown

Suppose that the area of the circle C is larger than the area of the triangle T = ¹ ⁄ ₂ cr . Let E stand for excess area. square in a circle so that all four corners of it lie on a circle. There are four segments between a square and a circle. If their total area G _{4 is} greater than E , we divide each arc in half, which turns the inscribed square into an octagon and forms eight segments with a smaller total gap, G ₈ . Continue the division until the total gap G _n becomes less than E. Now the area of the inscribed polygon P _n = C - G _n should be larger than the area of the triangle.

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

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

But this leads to a contradiction. To prove this, we draw the height from the center of the circle to the middle of the side of the polygon; its length h is less than the radius of the circle. Let each side of the polygon have length s , the sum of all sides will be ns , and this value is less than the circumference. The area of the polygon consists of n equal triangles of height h with base s , which gives ¹ ⁄ ₂ nhs . But h < r and ns < c , so that the area of the polygon must be less than the area of the triangle ¹ ⁄ ₂ cr , we get a contradiction.

No less

A circle with the described square and octagon. Clearance shown

Assume that the area of the circle is less than the area of the triangle. Let D be the difference in area. We describe the square around the circle, so that the midpoints of the sides lie on it. If the total gap between the square and the circle G _{4 is} greater than D , we cut off the corners by tangents, turning the square into an octagon and continue such cutoffs until the area of the gap becomes less than D. The area of the polygon P _n must be less than T.

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

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

This also leads to a contradiction. Each perpendicular drawn from the center of the circle to the middle of the side is a radius, i.e. has a length r . And since the sum of the sides is greater than the circumference, a polygon of n identical triangles will give an area greater than T. Again got a contradiction.

Thus, the area of the circle is exactly equal to the area of the triangle.

Reordering Proof

Circle area after regrouping

Rearrangement Animation

Following Sato Moshunu ^[4] and Leonardo da Vinci ^[5] , we can use the inscribed regular polygons in another way. Suppose we entered a hexagon . We cut the hexagon into six triangles, making sections through the center. Two opposite triangles contain common diameters. Now we move the triangles so that the radial sides become adjacent. Now a pair of triangles forms a parallelogram in which the sides of the hexagon form two opposite sides of length s . Two radial sides become lateral sides, and the height of the parallelogram is h (as in the proof of Archimedes). In fact, we can collect all the triangles into one large parallelogram, arranging the resulting parallelograms (from two triangles) in a row. The same will be true if we increase the number of parties. For a polygon with 2 n sides, the parallelogram will have a base ns and a height h . With an increase in the number of sides, the length of the base of the parallelogram increases, tending to half the circle, and the height tends to radius. In the limit, the parallelogram becomes a rectangle with a width π r and a height r .

**Approximations of the area of a circle of unit radius by a rearrangement of triangles.**
polygon
n	side	base	height	square
four	1.4142136	2,8284271	0.7071068	2,0000000
6	1,0000000	3,0000000	0.8660254	2,5980762
eight	0.7653669	3.0614675	0.9238795	2,8284271
ten	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
sixteen	0.3901806	3.1214452	0.9807853	3.0614675
96	0,0654382	3.1410320	0,9994646	3,1393502
∞	1 / ∞	π	one	π

Integration

Circle area by integration

Using the integrals, we can summarize the area of the circle, dividing it into concentric circles like a bulb . The area of the infinitely thin “layer” of radius t will be 2 π t dt , that is, the product of the circumference by the thickness of the layer. As a result, we obtain the elementary integral for a circle of radius r .

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

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

You can split a circle not into rings, but into triangles with an infinitely small base. The area of each such triangle is 1/2 * r * dt. Summing up (integrating) all the areas of these triangles, we obtain the circle formula:

{\begin{aligned}\mathrm {Area} (r)&{}=\int _{0}^{2\pi r}{\frac {1}{2}}r\,dt\\&{}=\left[{\frac {1}{2}}rt\right]_{t=0}^{2\pi r}\\&{}=\pi r^{2}.\end{aligned}}

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

Quick approximation

The calculations made by Archimedes were time-consuming and he settled on a polygon with 96 sides. The faster method uses the ideas of Snell (1621), later developed by Huygens (1654) ^[6] .

Archimedes Doubling Method

If a circle is given, let u _n be the perimeter of the inscribed regular n- gon and U _{n the} perimeter of the described regular n- gon. Then u _n and U _n are the lower and upper bounds on the circumference, which become more accurate with increasing n , and their average value ( u _n + U _n ) / 2 becomes a particularly good approximation of the circumference. To calculate u _n and U _n for large n , Archimedes derived the following formulas:

u_{2n}={\sqrt {U_{2n}u_{n}}}

{\ displaystyle u_ {2n} = {\ sqrt {U_ {2n} u_ {n}}}}

( geometric mean )

U_{2n}={\frac {2U_{n}u_{n}}{U_{n}+u_{n}}}

{\ displaystyle U_ {2n} = {\ frac {2U_ {n} u_ {n}} {U_ {n} + u_ {n}}}}

( harmonic mean ).

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

In modern notation, you can reproduce these calculations (and go further). For a unit circle, the inscribed hexagon has a perimeter u ₆ = 6, and the described hexagon has a perimeter U ₆ = 4√3. Double seven times, get

**Doubling Archimedes seven times;** **n = 6 × 2 ^k .**
k	n	u _n	U _n	( u _n + U _n ) / 4
0	6	6,0000000	6,9282032	3.2320508
one	12	6,2116571	6,4307806	3,1606094
2	24	6,2652572	6,3193199	3,1461443
3	48	6,2787004	6.2921724	3.1427182
four	96	6.2820639	6,2854292	3.1418733
five	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 length of the unit circle, which is 2 π , so that ( u _n + U _n ) / 4 approximates π )

The last row of the table contains ³⁵⁵ ⁄ ₁₁₃ - the best rational approximation , that is, there is no approximation of the best of this with a denominator of up to 113.
The number ³⁵⁵ ⁄ ₁₁₃ is an excellent approximation for π , there is no rational number closer to π with the denominator up to 16604. ^[7]

Snell-Huygens Improvement

Snell proposed (and Huygens proved) closer borders than Archimedes:

n{\frac {3\sin {\frac {\pi }{n}}}{2+\cos {\frac {\pi }{n}}}}<\pi <n[2\sin {\frac {\pi }{3n}}+\tan {\frac {\pi }{3n}}].

{\ displaystyle n {\ frac {3 \ sin {\ frac {\ pi} {n}}} {2+ \ cos {\ frac {\ pi} {n}}}} <\ pi <n [2 \ sin {\ frac {\ pi} {3n}} + \ tan {\ frac {\ pi} {3n}}].}

For n = 48, the formula gives an approximation better (about 3.14159292) than the Archimedes method for n = 768.

Archimedes Doubling Formula Development

Circle with similar triangles, described, inscribed and additional.

Let one side of the inscribed regular n- gon have length s _n and let the points A and B be its ends. Let A ′ be the point on the circle opposite A, so that A′A is a diameter, and A′AB is an inscribed triangle, based on this diameter. By the Thales theorem, this triangle is rectangular (angle B of the line). Let the length A′B be equal to c _n and this length will be called the complement s _n . Then c _n ² + s _n ² = (2 r ) ² . Let the point C divide the arc AB in half, and let C ′ be the opposite point of the circle C. Then the length of CA is s _{2 n} , the length of C′A is c _{2 n} , and C′CA is again a right triangle based on the diameter of C′C. Since C divides the arc AB in half, the diameter of C′C is perpendicular to the chord AB, which it intersects at, say, P. The triangle C′AP is then rectangular and similar to C′CA, since they have a common angle C ′. We get that 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 divides A′A in half, so that the triangle OAP is similar to A′AB and the length OP is half the length of A′B. As a result, we get

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

{\ 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 equality, the segment C′P is equal to the sum of C′O + OP, which is r + ¹ ⁄ ₂ c _n , and the segment C′C is the diameter and its length is 2 r . For a single circle, we get the famous formula for doubling Ludolph Van Zeilain

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

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

If we now construct the correct described n- gon with the ″ B ″ side parallel to AB, then OAB and OA ″ B ″ are similar with the similarity relation A ″ B ″: AB = OC: OP. Denote the described side by S _n , then the ratio turns into S _n : s _n = 1: ¹ ⁄ ₂ c _n . (We again use the fact that OP is equal to half A′B.) We get

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

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

Denote the perimeter of the inscribed polygon by u _n = ns _n , and the one described by U _n = nS _n . Combining the equalities, we obtain

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

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

so that

u_{2n}^{2}=u_{n}U_{2n}.

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

Got the geometric mean .

Can also be deduced

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

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

or

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

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

Received harmonic mean .

Random Cast Approximation

Area of a unit circle by Monte Carlo methods. After 900 casts we get 4 × ⁷⁰⁹ ⁄ ₉₀₀ = 3.15111 ...

If more effective methods are not available, you can resort to "throwing darts." This Monte Carlo method uses the fact that, in case of random throwing, the points evenly spread over the area of the square in which the circle is located, the number of hits in the circle approaches the ratio of the area of the circle to the area of the square. This method should be taken as the last opportunity to calculate the area of a circle (or figures of any shape), since it requires a huge number of tests to obtain acceptable accuracy. To obtain an accuracy of 10 ^{- n} , about 100 ⁿ random trials are necessary ^[8] .

Final regrouping

As we saw, breaking a disk into an infinite number of pieces we can then assemble a rectangle from them. An interesting fact was discovered relatively recently by Lackovic ^[9] that we can split a circle into a large, but finite number of pieces, and then rearrange them into a square of the same area. The very question of such a final partition is called “ Tarski Circle Squaring ”.

Generalizations

We can stretch the circle to an ellipse shape. Since this stretching is a linear transformation of flatness, it changes the area, but maintains the ratio of the areas. This fact can be used to calculate the area of an arbitrary ellipse, starting from the area of a circle.

Let the unit ellipse be described by a square with side 2. The transformation transforms the circle into an ellipse by compressing or stretching the horizontal and vertical diameters to the small and large axis of the ellipse. The square becomes the rectangle circled around the ellipse. The ratio of the area of the circle to the area of the square is π / 4, and the ratio of the area of the ellipse to the area of the rectangle will also be π / 4. If a and b are the lengths of the small and large axes of the ellipse. The area of the rectangle will be equal to ab , and then the area of the ellipse is π ab / 4.

We can extend similar techniques to large dimensions. For example, if we want to calculate the volume inside the sphere, and we know the formula for the area of the sphere, we can use a technique similar to the onion approach for the circle.

Triangle Method

Circle expanded triangle

The circle and the triangle have the same area.

This method is a modification of the proof using circles. Imagine unfolding concentric circles into segments, we get a right triangle with a height r and a base 2 π r (obtained from the outer circle of the circle).

Calculation of the area of the triangle will give the area of the circle:

Area = ½ * base * height = ½ * 2 π r * r = π r ² .

Notes

↑ James Stewart. Single variable calculus early transcendentals .. - 5th .. - Toronto ON: Brook / Cole, 2003. - S. 3. - ISBN 0-534-39330-6 .
↑ Thomas L. Heath. A Manual of Greek Mathematics. - Courier Dover Publications, 2003. - S. 121-132 . - ISBN 0-486-43231-9 . .
↑ Hill, George. Lectures on Geometry for Beginners , page 124 (1894).
↑ Smith, Mikami, 1914 .
↑ Beckmann, 1976 .
↑ Gerretsen, Verdenduin, 1983 .
↑ Not all best rational approximations come down to continued fractions!
↑ Thijsse, 2006 .
↑ Laczkovich, 1990 .

Literature

Archimedes translated by Thomas Heath . The Works of Archimedes. - Dover , c. 260 BCE, publication year 2002. - S. 91–93 . - ISBN 978-0-486-42084-4 .
Petr Beckmann. A History of Pi. - St. Martin's Griffin , 1976. - ISBN 978-0-312-38185-1 .
J. Gerretsen, P. Verdenduin. Fundamentals of Mathematics, Volume II: Geometry. - MIT Press , 1983. - S. 243–250 . - ISBN 978-0-262-52094-2 .
Miklós Laczkovich. Equidecomposability and discrepancy: A solution to Tarski's circle squaring problem // Journal für die reine und angewandte Mathematik . - 1990 .-- T. 404 . - S. 77–117 . (inaccessible link)
Serge Lang. Math! : Encounters with High School Students. - Springer-Verlag , 1985. - ISBN 978-0-387-96129-3 .
David Eugene Smith, Yoshio Mikami. A history of Japanese mathematics. - Chicago: Open Court Publishing , 1914. - S. 130-132 . - ISBN 978-0-87548-170-8 .
JMThijsse. Computational Physics. - Cambridge University Press, 2006 .-- S. 273 . - ISBN 978-0-521-57588-1 .

Links

Area of a Circle Calculator
Area enclosed by a circle (with interactive animation)
Science news on tarski problem

[1] James Stewart. Single variable calculus early transcendentals .. - 5th .. - Toronto ON: Brook / Cole, 2003. - S. 3. - ISBN 0-534-39330-6 .

[heath-2] Thomas L. Heath. A Manual of Greek Mathematics. - Courier Dover Publications, 2003. - S. 121-132 . - ISBN 0-486-43231-9 . .

[3] Hill, George. Lectures on Geometry for Beginners , page 124 (1894).

[_84476fb46deb21ad-4] Smith, Mikami, 1914 .

[_844037f9845c48b9-5] Beckmann, 1976 .

[_d101be672d92d153-6] Gerretsen, Verdenduin, 1983 .

[7] Not all best rational approximations come down to continued fractions!

[_6f90fd2133bf14af-8] Thijsse, 2006 .

[_9b4da67bdb1f782e-9] Laczkovich, 1990 .

Area of ​​a circle