Area of Circle/Proof 2

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 $\rd t$.

This article contains statements that are justified by handwavery. In particular: The fact that the above is a valid approximation needs to be established. You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding precise reasons why such statements hold. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{Handwaving}} from the code. If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page.

\(\ds A\)

\(=\)

\(\ds \int_0^r 2 \pi t \rd t\)

Perimeter of Circle

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds \pi r^2\)

$\blacksquare$