# Area of Circle

## Theorem

The area $A$ of a circle is given by:

- $A = \pi r^2$

where $r$ is the radius of the circle.

## Proof 1

From Equation of Circle:

- $x^2 + y^2 = r^2$

Thus $y = \pm \sqrt {r^2 - x^2}$.

It follows that from the geometric interpretation of the definite integral:

\(\ds A\) | \(=\) | \(\ds \int_{-r}^r \paren {\sqrt {r^2 - x^2} - \paren {-\sqrt {r^2 - x^2} } } \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_{-r}^r 2 \sqrt {r^2 - x^2} \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_{-r}^r 2 r \sqrt {1 - \frac {x^2} {r^2} } \rd x\) |

Let $x = r \sin \theta$ (note that we can do this because $-r \le x \le r$).

Thus $\theta = \map \arcsin {\dfrac x r}$ and $\rd x = r \cos \theta \rd \theta$.

\(\ds A\) | \(=\) | \(\ds \int_{\map \arcsin {\frac {-r} r} }^{\map \arcsin {\frac r r} } 2 r^2 \sqrt {1 - \frac {\paren {r \sin \theta}^2} {r^2} } \cos \theta \rd \theta\) | Integration by Substitution | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {\cos^2 \theta} \cos \theta \rd \theta\) | Sum of Squares of Sine and Cosine | |||||||||||

\(\ds \) | \(=\) | \(\ds r^2 \int_{-\frac \pi 2}^{\frac \pi 2} 2 \cos^2 \theta \rd \theta\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r^2 \int_{-\frac \pi 2}^{\frac \pi 2} \paren {1 + \map \cos {2 \theta} } \rd \theta\) | Double Angle Formula for Cosine: Corollary 1 | |||||||||||

\(\ds \) | \(=\) | \(\ds r^2 \intlimits {\theta + \frac 1 2 \map \sin {2 \theta} } {-\frac \pi 2} {\frac \pi 2}\) | from Integration of Constant and Primitive of Cosine Function | |||||||||||

\(\ds \) | \(=\) | \(\ds r^2 \paren {\frac \pi 2 + \frac 1 2 \map \sin {2 \cdot \frac {-\pi} 2} - \frac {-\pi} 2 - \frac 1 2 \map \sin {2 \cdot \frac \pi 2} }\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds r^2 \paren {2 \cdot \frac \pi 2 + 2 \cdot \frac 1 2 \cdot 0}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \pi r^2\) |

$\blacksquare$

## Proof 2

Proof by shell integration:

The circle can be divided into a set of infinitesimally thin rings, each of which has area $2 \pi t \rd t$, since the ring has length $2 \pi t$ and thickness $\rd t$.

\(\ds A\) | \(=\) | \(\ds \int_0^r 2 \pi t \rd t\) | Perimeter of Circle | |||||||||||

\(\ds \) | \(=\) | \(\ds \bigintlimits {\pi t^2} 0 r\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \pi r^2\) |

$\blacksquare$

## Proof 3

Refer to the figure.

Construct a circle with radius r and circumference $c$, where its area is denoted by $C$.

Construct a triangle with height r and base $c$, where its area is denoted by $T$.

### Lemma 1: $T = \pi r^2$

\(\ds T\) | \(=\) | \(\ds \frac {r c} 2\) | Area of Triangle in Terms of Side and Altitude | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {r 2 \pi r} 2\) | Perimeter of Circle | |||||||||||

\(\ds \) | \(=\) | \(\ds \pi r^2\) |

$\Box$

### Lemma 2: $T \ge C$

Aiming for a contradiction, suppose $T < C$.

It should be possible to construct a regular polygon with area $P$, where $T < P < C$.

For any given regular polygon:

- $P = \dfrac {h q} 2$

where:

- $q$ is the perimeter of the polygon
- $h$ is the height of any given triangular part of it
- $P$ is the area.

On one hand:

- $P > T \implies \dfrac {h q} 2 > \dfrac {r c} 2$

On the other hand:

- $0 < h < r \land 0 < q < c \implies \dfrac {h q} 2 < \dfrac {r c} 2$

Hence a contradiction is obtained.

Hence $\neg T < C$.

Hence $T \ge C$.

$\Box$

### Lemma 3: $T \le C$

Aiming for a contradiction, suppose $T > C$.

It should be possible to construct a regular polygon with area $P$, where $C < P < T$.

From Area of Polygon by Inradius and Perimeter:

- $P = \dfrac {h q} 2$

where:

- $q$ is the perimeter of the regular polygon
- $h$ is the inradius of the regular polygon
- $P$ is the area.

as each triangle has the base $B = \dfrac q n$ and area $A = \dfrac {h q} {2 n}$ and with $n$ triangles we get $P = \dfrac {h q} 2$

On one hand:

- $P < T \implies \dfrac {h q} 2 < \dfrac {r c} 2$

On the other hand:

- $0 < h = r \land 0 < c < q \implies \dfrac {h q} 2 > \dfrac {r c} 2$

Hence a contradiction is obtained.

Hence $\neg T > C$.

Hence $T \le C$.

$\Box$

### Final Proof

- $T \ge C$ (from Lemma 2)
- $T \le C$ (from Lemma 3)
- $\therefore T \mathop = C$
- $\therefore C \mathop = T \mathop = \pi r^2$ (from Lemma 1)

$\blacksquare$

## Proof 4

Expressing the area in polar coordinates:

\(\ds \iint \rd A\) | \(=\) | \(\ds \int_0^r \int_0^{2 \pi} t \rd t \rd \theta\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \intlimits {\int_0^r t \theta} 0 {2 \pi} \rd t\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \int_0^r 2 \pi t \rd t\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \pi \paren {\intlimits {\frac 1 2 t^2} 0 r}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 2 \pi \paren {\frac 1 2 r^2}\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \pi r^2\) |

$\blacksquare$

## Proof 5

Let the circle of radius $r$ be divided into many sectors:

If they are made small enough, they can be approximated to triangles whose heights are all $r$.

Let the bases of these triangles be denoted:

- $b_1, b_2, b_3, \ldots$

From Area of Triangle in Terms of Side and Altitude, their areas are:

- $\dfrac {r b_1} 2, \dfrac {r b_2} 2, \dfrac {r b_3} 2, \ldots$

The area $\mathcal A$ of the circle is given by the sum of the areas of each of these triangles:

\(\ds \mathcal A\) | \(=\) | \(\ds \dfrac {r b_1} 2 + \dfrac {r b_2} 2 + \dfrac {r b_3} 2 + \cdots\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac r 2 \left({b_1 + b_2 + b_3 + \cdots}\right)\) |

But $b_1 + b_2 + b_3 + \cdots$ is the length of the circumference of the circle.

From Perimeter of Circle:

- $b_1 + b_2 + b_3 + \cdots = 2 \pi r$

Hence:

\(\ds \mathcal A\) | \(=\) | \(\ds \dfrac r 2 \left({b_1 + b_2 + b_3 + \cdots}\right)\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \dfrac r 2 \left({2 \pi r}\right)\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \pi r^2\) |

It needs to be noted that this proof is intuitive and non-rigorous.

$\blacksquare$

## Proof 6

From Equation of Circle:

- $x^2 + y^2 = r^2$

Let $A$ be the area of the circle whose equation is given by $x^2 + y^2 = r^2$.

We have that:

- $y = \pm \sqrt {r^2 - x^2}$

For the upper half of the circle:

- $y = +\sqrt {r^2 - x^2}$

Thus for the right hand half of the upper half of the circle:

\(\ds \frac A 4\) | \(=\) | \(\ds \int_0^r \sqrt {r^2 - x^2} \rd x\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\pi r^2} 4\) | Definite Integral from $0$ to $r$ of $\sqrt {r^2 - x^2}$ |

Hence the result.

$\blacksquare$

## Proof 7

By the method of exhaustion:

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 4$: Geometric Formulas: $4.11$ - 1986: David Wells:
*Curious and Interesting Numbers*... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41972 \ldots$ - 1997: David Wells:
*Curious and Interesting Numbers*(2nd ed.) ... (previous) ... (next): $3 \cdotp 14159 \, 26535 \, 89793 \, 23846 \, 26433 \, 83279 \, 50288 \, 41971 \ldots$ - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Entry:**circle** - 2014: Christopher Clapham and James Nicholson:
*The Concise Oxford Dictionary of Mathematics*(5th ed.) ... (previous) ... (next): Appendix $1$: Areas and volumes