Area of Circle/Kepler's Proof
Theorem
The area $A$ of a circle is given by:
- $A = \pi r^2$
where $r$ is the radius of the circle.
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.
$\blacksquare$
Historical Note
This was the method used by Johannes Kepler to calculate the area of a circle when he was working on his Second Law of Planetary Motion.
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {A}.10$: Kepler ($\text {1571}$ – $\text {1630}$)