# Area of Sector

## Theorem

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

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

Then the area $\AA$ of sector $BAC$ is given by:

- $\AA = \dfrac {r^2 \theta} 2$

where:

## Proof 1

Let $\CC$ 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 $\CC$.

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

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

We have that $\CC$ is centered at the origin.

From Equation of Circle center Origin, the equation of $\CC$ 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:

\(\ds \sin \theta\) | \(=\) | \(\ds \frac {\size {BD} } {\size {AB} }\) | Definition of Sine of Angle | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {x_0} r\) | Definition of $x_0$ and $r$ | |||||||||||

\(\ds \leadsto \ \ \) | \(\ds \map \arcsin {\sin \theta}\) | \(=\) | \(\ds \map \arcsin {\frac {x_0} r}\) | taking $\arcsin$ of both sides | ||||||||||

\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \theta\) | \(=\) | \(\ds \map \arcsin {\frac {x_0} r}\) | 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 $AB$.

\(\ds \Delta y\) | \(=\) | \(\ds \sqrt {r^2 - {x_0}^2} - 0\) | Difference between $y$ values of $A$ and $B$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \sqrt {r^2 - {x_0}^2}\) | ||||||||||||

\(\ds \Delta x\) | \(=\) | \(\ds x_0 - 0\) | Difference between $x$ values of $A$ and $B$ | |||||||||||

\(\ds \) | \(=\) | \(\ds x_0\) | ||||||||||||

\(\ds \leadsto \ \ \) | \(\ds m\) | \(=\) | \(\ds \frac {\Delta y} {\Delta x}\) | Definition of Slope of Straight Line | ||||||||||

\(\ds \) | \(=\) | \(\ds \frac {\sqrt {r^2 - {x_0}^2} } {x_0}\) | ||||||||||||

\(\text {(2)}: \quad\) | \(\ds y\) | \(=\) | \(\ds \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x\) | Equation of Straight Line in Plane in Slope-Intercept Form: $y$ intercept is $0$ |

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

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

Let $\AA$ be the area of $K$.

Then:

\(\ds \AA\) | \(=\) | \(\ds \int_0^{x_0} \paren {\sqrt {r^2 - x^2} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x} \rd x\) | integrating between two curves to find area | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_0^{x_0} \sqrt {r^2 - x^2} \rd x - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \int_0^{x_0} x \rd x\) | Linear Combination of Integrals | |||||||||||

\(\ds \) | \(=\) | \(\ds \int_0^{x_0} \sqrt {r^2 - x^2} \rd x - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \sqbrk {\frac {x^2} 2}_0^{x_0}\) | Primitive of Power | |||||||||||

\(\ds \) | \(=\) | \(\ds \sqbrk {\frac {x \sqrt {r^2 - x^2} } 2 + \frac {r^2} 2 \arcsin \frac x r}_0^{x_0} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \sqbrk {\frac {x^2} 2}_0^{x_0}\) | Primitive of $\sqrt {a^2 - x^2}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2 + \frac {r^2} 2 \arcsin \frac {x_0} r - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} \frac { {x_0}^2} 2\) | evaluating over range $\closedint 0 {x_0}$ | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2 + \frac {r^2} 2 \arcsin \frac {x_0} r - \frac {x_0 \sqrt {r^2 - {x_0}^2} } 2\) | simplifying | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {r^2} 2 \arcsin \frac {x_0} r\) | cancelling | |||||||||||

\(\ds \) | \(=\) | \(\ds \frac {r^2 \theta} 2\) | from $(1)$ |

$\blacksquare$

## Proof 2

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

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

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

Hence the result.

$\blacksquare$

## Sources

- 1968: Murray R. Spiegel:
*Mathematical Handbook of Formulas and Tables*... (previous) ... (next): $\S 4$: Geometric Formulas: $4.13$

- Weisstein, Eric W. "Circular Sector." From
*MathWorld*--A Wolfram Web Resource. http://mathworld.wolfram.com/CircularSector.html