# Area of Circle/Proof 2

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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

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 | ||||||||||

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

\(\displaystyle \) | \(=\) | \(\displaystyle \pi r^2\) |

$\blacksquare$