Question

How do you prove that the circumference of a circle is $2\mathrm{\pi }r$?

Answer 1

Derive the expression for circumference of a circle:

Let $r$ be the radius a circle.

Let us slice the circumference into small sections and consider a small angle of of the circle subtended by two of its radii. This angle is,
$d\theta =\frac{dl}{r}$
where $dl$ is a small part of the circumference of the circle.

Integrating,

$\begin{array}{cc}\Rightarrow & \int d\theta =\int \frac{dl}{r}\\ \Rightarrow & \int d\theta =\frac{1}{r}\int dl\dots \left(1\right)\end{array}$

$\int d\theta$ is adding up all the small angles of the circle which is the total angle of a circle. And we know that the total angle of a circle is $2\mathrm{\pi }$. i.e.,
$\int d\theta =2\mathrm{\pi }$

$\int dl$ is adding up all the small parts of the circumference which results in the circumference of the circle. i.e.,
$\int dl=C$

Substituting the above values into $\left(1\right)$,

$\begin{array}{cc}\Rightarrow & 2\mathrm{\pi }=\frac{C}{\mathrm{r}}\\ \therefore & C=2\mathrm{\pi }r\end{array}$

Hence proved