Area of Circle/Proof 2

From ProofWiki
Jump to navigation Jump to search

Theorem

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

$A = \pi r^2$

where $r$ is the radius of the circle.


Proof

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

$\blacksquare$