Area of Sector

Theorem

Let $\mathcal C = ABC$ be a circle whose center is $A$ and with radii $AB$ and $AC$.

Let $BAC$ be the sector of $\mathcal C$ whose angle between $AB$ and $AC$ is $\theta$. Then the area $\mathcal A$ of sector $BAC$ is given by:

$\mathcal A = \dfrac {r^2 \theta} 2$

where:

$r = AB$ is the length of the radius of the circle
$\theta$ is measured in radians.

Proof 1

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

Let $B$ and $C$ be arbitrary points on the circumference of $\mathcal C$.

Let $C$ be positioned at $\tuple {0, r}$ so that $AC$ coincides with the $y$-axis.

Let $BAC$ be the sector of $\mathcal C$ whose angle between $AB$ and $AC$ is $\theta$. We have that $\mathcal C$ is centered at the origin.

From Equation of Circle: Corollary 2, the equation of $\mathcal C$ is given by:

$r^2 = x^2 + y^2$

Let $B$ have co-ordinates:

$\paren {x_0, y_0} = \paren {x_0, \sqrt {r^2 - {x_0}^2} }$

Triangle $\triangle ADB$ is a right triangle whose leg $DB$ is opposite $\theta$, and whose hypotenuse is $AB$.

Hence:

 $\displaystyle \sin \theta$ $=$ $\displaystyle \frac {\size {BD} } {\size {AB} }$ Definition of Sine $\displaystyle$ $=$ $\displaystyle \frac {x_0} r$ Definition of $x_0$ and $r$ $\displaystyle \leadsto \ \$ $\displaystyle \map \arcsin {\sin \theta}$ $=$ $\displaystyle \map \arcsin {\frac {x_0} r}$ taking $\arcsin$ of both sides $(1):\quad$ $\displaystyle \leadsto \ \$ $\displaystyle \theta$ $=$ $\displaystyle \map \arcsin {\frac {x_0} r}$ Definition of Arcsine

Consider the line $AB$, which by definition passes through $A = \paren {0, 0}$ and $B = \paren {x_0, \sqrt {r^2 - {x_0}^2} }$.

Let:

$\Delta y$ be the change in $y$ between $A$ and $B$
$\Delta x$ be the change in $x$
$m$ be the slope of $AB$.
 $\displaystyle \Delta y$ $=$ $\displaystyle \sqrt {r^2 - {x_0}^2} - 0$ Difference between $y$ values of $A$ and $B$ $\displaystyle$ $=$ $\displaystyle \sqrt {r^2 - {x_0}^2}$ $\displaystyle \Delta x$ $=$ $\displaystyle x_0 - 0$ Difference between $x$ values of $A$ and $B$ $\displaystyle$ $=$ $\displaystyle x_0$ $\displaystyle \leadsto \ \$ $\displaystyle m$ $=$ $\displaystyle \frac {\Delta y} {\Delta x}$ Definition of Slope of Straight Line $\displaystyle$ $=$ $\displaystyle \frac {\sqrt {r^2 - {x_0}^2} } {x_0}$ $(2):\quad$ $\displaystyle y$ $=$ $\displaystyle \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x$ Equation of Straight Line in Plane in Slope-Intercept Form: $y$ intercept is $0$

Let $K$ be the region between $\mathcal C$, $AC$ and $AB$, from $0$ to $x_0$, coloured orange in the diagram above.

This is the sector of $\mathcal C$ whose area we are to find.

Let $\mathcal A$ be the area of $K$.

Then:

 $\displaystyle \mathcal A$ $=$ $\displaystyle \int_0^{x_0} \paren {\sqrt {r^2 - x^2} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x} \rd x$ integrating between two curves to find area $\displaystyle$ $=$ $\displaystyle \int_0^{x_0} \sqrt {r^2 - x^2} \rd x - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \int_0^{x_0} x \rd x$ Linear Combination of Integrals $\displaystyle$ $=$ $\displaystyle \int_0^{x_0} \sqrt {r^2 - x^2} \rd x - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \sqbrk {\frac {x^2} 2}_0^{x_0}$ Primitive of Power $\displaystyle$ $=$ $\displaystyle \sqbrk {\frac {x \sqrt {r^2 - x^2} } 2 + \frac {r^2} 2 \arcsin \frac x r}_0^{x_0} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \sqbrk {\frac {x^2} 2}_0^{x_0}$ Primitive of $\sqrt {a^2 - x^2}$ $\displaystyle$ $=$ $\displaystyle \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2 + \frac {r^2} 2 \arcsin \frac {x_0} r - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \frac { {x_0}^2} 2$ evaluating over range $\closedint 0 {x_0}$ $\displaystyle$ $=$ $\displaystyle \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2 + \frac {r^2} 2 \arcsin \frac {x_0} r - \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2$ simplifying $\displaystyle$ $=$ $\displaystyle \frac {r^2} 2 \arcsin \frac {x_0} r$ cancelling $\displaystyle$ $=$ $\displaystyle \frac {r^2 \theta} 2$ from $(1)$

$\blacksquare$

Proof 2

From Area of Circle, the area of $\mathcal C$ is $\pi r^2$.

From Full Angle measures $2 \pi$ Radians, the angle within $\mathcal C$ is $2 \pi$.

The fraction of the area of $\mathcal C$ within the sector $BAC$ is therefore $\pi r^2 \times \dfrac \theta {2 \pi}$.

Hence the result.

$\blacksquare$