Question

Find the area of the region in the first quadrant enclosed by $x-\text{axis}$, line $x=\sqrt{3}y$ and the circle $x^2+y^2=4$.

Answer 1

Given equation

$x=\sqrt{3}y$

$x^2+y^2=4$

$\therefore$ Radius $r=2$ Centre $=(0,0)$

Draw diagram
$\Rightarrow (\sqrt{3}y{)}^2+y^2=4$

$\Rightarrow 3y^2+y^2=4$

$\Rightarrow 4y^2=4$

$\Rightarrow y^2=1$

$\therefore y=\pm 1$

When $y=1$	When $y=-1$
$x=\sqrt{3}y$	$x=\sqrt{3}y$
$x=\sqrt{3}$	$x=-\sqrt{3}$

$\therefore C$ is $\sqrt{3},1$

$x^2+y^2=4$

$\Rightarrow y^2=4-x^2$

$\Rightarrow y=\pm \sqrt{4-x^2}$

$\therefore y=\sqrt{4-x^2}$

$\text{Area of}OACO=\text{Area}OCXO+\text{Area}XCAX={\int }_0^{\sqrt{3}}ydx+{\int }_{\sqrt{3}}^{2}ydx$

$={\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

$=\frac{1}{\sqrt{3}}{\int }_0^{\sqrt{3}}xdx+{\int }_{\sqrt{3}}^2\sqrt{(2{)}^2-x^2}dx$

$=\frac{1}{\sqrt{3}}{\left[\frac{x^2}{2}\right]}_0^{\sqrt{3}}+{[\frac{x}{2}\sqrt{(2{)}^2-x^2}+\frac{(2{)}^2}{2}{\sin }^{-1}\frac{x}{a}]}_{\sqrt{3}}^2$

$=\frac{3}{2\sqrt{3}}+2{\sin }^{-1}\left(1\right)-\frac{\sqrt{3}}{2}-2{\sin }^{-1}\frac{\sqrt{3}}{2}$

$=\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}+2[\frac{\pi }{2}-\frac{\pi }{3}]=2\left[\frac{\pi }{6}\right]$

$=\frac{\pi }{3}$

Final answer:
Therefore, required area $=\frac{\pi }{3}\text{square units}$.