Area of Sector

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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 $\theta$.

Sector.png


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

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

where:

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


Proof 1

Let $\CC$ 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 $\theta$.


Area-of-Sector.png


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:

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

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 Real Arcsine


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


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 Definite 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


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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

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