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 eqn 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:
 * $\displaystyle A = \int_{-r}^r \left[ \sqrt{r^2 - x^2} - (-\sqrt{r^2 - x^2})\right] dx$

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

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

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