Area of Circle

Theorem
The area $A$ of a circle is given by the formula $A=\pi r^2$, where $r$ is the radius of the circle.

Proof
We start with the equation of a circle:
 * $x^2 + y^2 = r^2$

Thus $y = \pm \sqrt{r^2 - x^2}$, so from the geometric interpretation of the definite integral:

Let $x = r \sin \theta$ (note that we can do this because $-r \le x \le r$).

Thus $\theta = \arcsin \left({\dfrac x r}\right)$ and $\mathrm d x = r \cos \theta \ \mathrm d \theta$.

Proof by Shell Integration
The circle can be divided into a set of infinitesimally thin rings, each of which has area $2 \pi t \ \mathrm dt$, since the ring has length $2 \pi t$ and thickness $\mathrm d t$.