Talk:Area of Sector

This seems to be assuming that the area of the sector is proportional to the length of the arc which is the measure of the angle. This looks quite plausible from the picture, but I do not see any proof here.

The way to prove this is to calculate both by doing different integrals and showing the results are in proportion. (They are both proportional the $\arcsin\theta$ ). --Pelliott (talk) 18:26, 13 January 2019 (EST)


 * Good call. Go to it. --prime mover (talk) 04:54, 13 January 2019 (EST)

I propose replacing this whole page

New version reads as follows (What do you think):

Theorem
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 $\left(0,0\right)$. 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 $\triangle y$ be the change in $y$ between these two points, $\triangle x$ 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.