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

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

### Lemma $1$

$T = \pi r^2$

$\Box$

### Lemma $2$

$T \ge C$

$\Box$

### Lemma $3$

$T \le C$

$\Box$

### Final Proof

From Lemma $2$:

$T \ge C$

From Lemma $3$:

$T \le C$

Therefore:

$T \mathop = C$

and so from Lemma $1$:

$C \mathop = T \mathop = \pi r^2$

$\blacksquare$

## Proof 4

Expressing the area in polar coordinates:

 $\ds \iint \rd A$ $=$ $\ds \int_0^r \int_0^{2 \pi} t \rd t \rd \theta$ $\ds$ $=$ $\ds \intlimits {\int_0^r t \theta} 0 {2 \pi} \rd t$ $\ds$ $=$ $\ds \int_0^r 2 \pi t \rd t$ $\ds$ $=$ $\ds 2 \pi \paren {\intlimits {\frac 1 2 t^2} 0 r}$ $\ds$ $=$ $\ds 2 \pi \paren {\frac 1 2 r^2}$ $\ds$ $=$ $\ds \pi r^2$

$\blacksquare$

## Kepler's Proof

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 $\AA$ of the circle is given by the sum of the areas of each of these triangles:

 $\ds \AA$ $=$ $\ds \dfrac {r b_1} 2 + \dfrac {r b_2} 2 + \dfrac {r b_3} 2 + \cdots$ $\ds$ $=$ $\ds \dfrac r 2 \paren {b_1 + b_2 + b_3 + \cdots}$

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:

 $\ds \AA$ $=$ $\ds \dfrac r 2 \paren {b_1 + b_2 + b_3 + \cdots}$ $\ds$ $=$ $\ds \dfrac r 2 \paren {2 \pi r}$ $\ds$ $=$ $\ds \pi r^2$

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: 