Area of Circle

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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:

\(\displaystyle A\) \(=\) \(\displaystyle \int_{-r}^r \left({\sqrt {r^2 - x^2} - \left({-\sqrt {r^2 - x^2} }\right)}\right) \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{-r}^r 2 \sqrt {r^2 - x^2} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{-r}^r 2 r \sqrt {1 - \frac {x^2} {r^2} } \rd x\)

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

Thus $\theta = \arcsin \left({\dfrac x r}\right)$ and $\rd x = r \cos \theta \rd \theta$.

\(\displaystyle A\) \(=\) \(\displaystyle \int_{\arcsin \left({\frac {-r} r}\right)}^{\arcsin \left({\frac r r}\right)} 2 r^2 \sqrt {1 - \frac {\left({r \sin \theta}\right)^2} {r^2} } \cos \theta \rd \theta\) Integration by Substitution
\(\displaystyle \) \(=\) \(\displaystyle \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {1 - \sin^2 \theta} \cos \theta \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle \int_{-\frac \pi 2}^{\frac \pi 2} 2 r^2 \sqrt {\cos^2 \theta} \cos \theta \rd \theta\) Sum of Squares of Sine and Cosine
\(\displaystyle \) \(=\) \(\displaystyle r^2 \int_{-\frac \pi 2}^{\frac \pi 2} 2 \cos^2 \theta \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle r^2 \int_{-\frac \pi 2}^{\frac \pi 2} \left({1 + \cos \left({2 \theta}\right)}\right) \rd \theta\) Double Angle Formula for Cosine: Corollary 1
\(\displaystyle \) \(=\) \(\displaystyle r^2 \left[{\theta + \frac 1 2 \sin \left({2 \theta}\right)}\right]_{-\frac \pi 2}^{\frac \pi 2}\) from Integration of Constant and Primitive of Cosine Function
\(\displaystyle \) \(=\) \(\displaystyle r^2 \left[{\frac \pi 2 + \frac 1 2 \sin \left({2 \cdot \frac {-\pi} 2}\right) - \frac {-\pi} 2 - \frac 1 2 \sin \left({2 \cdot \frac \pi 2}\right)}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle r^2 \left[{2 \cdot \frac \pi 2 + 2 \cdot \frac 1 2 \cdot 0}\right]\)
\(\displaystyle \) \(=\) \(\displaystyle \pi r^2\)

$\blacksquare$


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 $\rd t$.


\(\displaystyle A\) \(=\) \(\displaystyle \int_0^r 2 \pi t \rd t\) Perimeter of Circle
\(\displaystyle \) \(=\) \(\displaystyle \bigintlimits {\pi t^2} 0 r\)
\(\displaystyle \) \(=\) \(\displaystyle \pi r^2\)

$\blacksquare$


Proof 3

Area Equal.jpg

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

\(\displaystyle T\) \(=\) \(\displaystyle \frac{rc}2\) Area of Triangle in Terms of Side and Altitude
\(\displaystyle \) \(=\) \(\displaystyle \frac{r 2 \pi r} 2\) Perimeter of Circle
\(\displaystyle \) \(=\) \(\displaystyle \pi r^2\)

$\Box$


Lemma 2: $T \ge C$

Area Smaller.jpg

This will be proven by contradiction.

Assume $T < C$.

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

For any given regular polygon:

$P = \dfrac {hq} 2$

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:

$P > T \implies \dfrac {hq} 2 > \dfrac {rc} 2$

On the other hand:

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

Hence a contradiction is obtained.

Hence $\neg T < C$.

Hence $T \ge C$.

$\Box$


Lemma 3: $T \le C$

Area Greater.jpg

This will be proven by contradiction.

Assume $T > C$.

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

From Area of Polygon by Inradius and Perimeter:

$P = \dfrac {hq} 2$

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 $A = \dfrac {hq}{2n}$ and with $n$ triangles we get $P = \dfrac {hq} 2$

On one hand:

$P < T \implies \dfrac {hq} 2 < \dfrac {rc} 2$

On the other hand:

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

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:

\(\displaystyle \iint \rd A\) \(=\) \(\displaystyle \int_0^r \int_0^{2 \pi} t \rd t \rd \theta\)
\(\displaystyle \) \(=\) \(\displaystyle \left.{\int_0^r t \theta}\right\vert_0^{2 \pi} \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \int_0^r 2 \pi t \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \paren {\left.{\frac 1 2 t^2}\right\vert_0^r}\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \pi \paren {\frac 1 2 r^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \pi r^2\)

$\blacksquare$


Proof 5

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

AreaOfCircleProof5.png

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:

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

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:

\(\displaystyle \mathcal A\) \(=\) \(\displaystyle \dfrac r 2 \left({b_1 + b_2 + b_3 + \cdots}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \dfrac r 2 \left({2 \pi r}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \pi r^2\)

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:

\(\displaystyle \frac A 4\) \(=\) \(\displaystyle \int_0^r \sqrt {r^2 - x^2} \rd x\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi r^2} 4\) Definite Integral from $0$ to $r$ of $\sqrt {r^2 - x^2}$

Hence the result.

$\blacksquare$


Proof 7

By the method of exhaustion:

AreaOfCircleMethodOfExhaustion.png




Sources

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