Area of Circle/Proof 2

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

$A = \pi r^2$

where $r$ is the radius of the circle.


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 $\rd t$.

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