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 $\theta$. <!-- :660px --> :660px We have that $\CC$ is centered at the origin. From Equation of Circle: Corollary 2, 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: {{begin-eqn}} {{eqn | l = \sin \theta       | r = \frac {\size {BD} } {\size {AB} }       | c = {{Defof|Sine of Angle}} }} {{eqn | r = \frac {x_0} r       | c = Definition of $x_0$ and $r$ }} {{eqn | ll= \leadsto       | l = \map \arcsin {\sin \theta}       | r = \map \arcsin {\frac {x_0} r}       | c = taking $\arcsin$ of both sides }} {{eqn | ll= \leadsto       | l = \theta       | r = \map \arcsin {\frac {x_0} r}       | c = {{Defof|Arcsine}}       | n = 1 }} {{end-eqn}} 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 $AB$. {{begin-eqn}} {{eqn | l = \Delta y       | r = \sqrt {r^2 - {x_0}^2} - 0       | c = Difference between $y$ values of $A$ and $B$ }} {{eqn | r = \sqrt {r^2 - {x_0}^2} }} {{eqn | l = \Delta x       | r = x_0 - 0       | c = Difference between $x$ values of $A$ and $B$ }} {{eqn | r = x_0 }} {{eqn | ll= \leadsto       | l = m       | r = \frac {\Delta y} {\Delta x}       | c = {{Defof|Slope of Straight Line}} }} {{eqn | r = \frac {\sqrt {r^2 - {x_0}^2} } {x_0} }} {{eqn | l = y       | r = \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x       | c = Equation of Straight Line in Plane in Slope-Intercept Form: $y$ intercept is $0$       | n = 2 }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \AA       | r = \int_0^{x_0} \paren {\sqrt {r^2 - x^2} - \frac {\sqrt {r^2 - {x_0}^2} } {x_0} x} \rd x       | c = integrating between two curves to find area }} {{eqn | r = \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       | c = Linear Combination of Integrals }} {{eqn | r = \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}       | c = Primitive of Power }} {{eqn | r = \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}       | c = Primitive of $\sqrt {a^2 - x^2}$ }} {{eqn | r = \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       | c = evaluating over range $\closedint 0 {x_0}$ }} {{eqn | r = \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       | c = simplifying }} {{eqn | r = \frac {r^2} 2 \arcsin \frac {x_0} r       | c = cancelling }} {{eqn | r = \frac {r^2 \theta} 2       | c = from $(1)$ }} {{end-eqn}} {{qed}}