Area of Sector/Proof 1
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
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:
- $\tuple {x_0, y_0} = \tuple {x_0, \sqrt {r^2 - {x_0}^2} }$
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 $(1)$ |
$\blacksquare$