Question

Find the area bounded by the circle $x^2+y^2=16$ and the line $\sqrt{3y}=x$ in the first quadrant, using integration.

Answer 1

Equation of the circle is $x^2+y^2=4$ where radius$r=2$units

Let $AB$ represent the line $x=\sqrt{3}y$

Point $C$ is the intersectio of line and circle.

Put $x=\sqrt{3}y$ in $x^2+y^2=4$ we get

$3y^2+y^2=4$ or $4y^2=4$ or $y=\pm 1$

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

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

As point $C$ is in first quadrant $C$ is $(\sqrt{3},1)$

Now,area of $△OAC=$Area$△OCX+$Area$△XCA$

$={\int }_0^{\sqrt{3}}{y}_{1}dx+{\int }_{\sqrt{3}}^2{y}_{2}dx$

Put $x=\sqrt{3}y\Rightarrow y=\frac{x}{\sqrt{3}}$ So,$y_1=\frac{x}{\sqrt{3}}$

$x^2+y^2=4\Rightarrow y^2=4-x^2$ or $y_2=\sqrt{4-x^2}$

$\therefore$ Area of $△OAC={\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx+{\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$ .......$\left(1\right)$

Let $I_1={\int }_0^{\sqrt{3}}\frac{x}{\sqrt{3}}dx$

$=\frac{1}{\sqrt{3}}{\left[\frac{x^2}{2}\right]}_0^{\sqrt{3}}$

$=\frac{1}{2\sqrt{3}}(3-0)$

$=\frac{\sqrt{3}}{2}$

Now, $I_2={\int }_{\sqrt{3}}^2\sqrt{4-x^2}dx$

$={[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\frac{x}{2}]}_{\sqrt{3}}^2$ since $\int \sqrt{a^2-x^2}dx=\frac{1}{2}x\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}+c$

$=\frac{1}{2}\left(2\right)\sqrt{4-4}-\frac{1}{2}\sqrt{3}\sqrt{4-3}+\frac{4}{2}{\sin }^{-1}\frac{2}{2}-\frac{4}{2}{\sin }^{-1}\frac{\sqrt{3}}{2}$

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

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

$=\frac{-\sqrt{3}}{2}+\frac{\pi }{3}$

Putting the value of $I_1$ and $I_2$ in $\left(1\right)$

Area of $△OAC=\frac{\sqrt{3}}{2}-\frac{-\sqrt{3}}{2}+\frac{\pi }{3}=\frac{\pi }{3}$