Area of Sector/Proof 1

Proof
Let $\mathcal C = ABC$ be a circle whose center is $A$ at the origin and with radii $AB$ and $AC$ with radius $r$.

Let $B$ be an arbitrary point on this circle.

Let $BAC$ be the sector of $\mathcal C$ whose angle between $AB$ and $AC$ is $\theta$.


 * sector-C.svg

The circle is centered at the origin with co-ordinates $\tuple {0, 0}$.

The equation of the circle is $r^2 = x^2 + y^2$.

Let the arbitrary point $B$ have co-ordinates $\left(x_{0},y_{0}\right)=\left(x_{0},\sqrt{r^{2}-x_{0}^{2}}\right)$.

Triangle $\triangle ADB$ is a right triangle with leg $\overline{DB}$, opposite $\theta$, and hypotenuse $\overline{AB}$.

Consider the line through $A=\left(0,0\right)$, and $B=\left(x_{0},\sqrt{r^{2}-x_{0}^{2} }\right)$.

Let $\Delta y$ be the change in $y$ between these two points, $\Delta x$ be the change in $x$, and $m$ be the slope of the line.

Let $K$ be the red area between the circle and the line, from $0$ to $x_{0}$.

This is given by the following integral.