Area of Circle/Proof 1

Proof
From Equation of Circle:
 * $x^2 + y^2 = r^2$

Thus $y = \pm \sqrt{r^2 - x^2}$.

It follows that 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 $\mathrm d x = r \cos \theta \ \mathrm d \theta$.