Question

The area bounded by the circle $x^2+y^2=4$ and the line $x=y\sqrt{3}$ in the first quadrant (in sq units) is

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. None of these

Answer 1

Answer: (C)

We have, circle $x^2+y^2=4...(i)$
and line, $x=y\sqrt{3}$
$\Rightarrow y=\frac{x}{\sqrt{3}}...(ii)$

On putting the value of $y$ from Eq, $(ii)$ to Eq. $(i)$, we get
$x^2+(\frac{x}{\sqrt{3}}{)}^2=4$
$\Rightarrow x^2+\frac{x^2}{3}=4$
$\Rightarrow 3x^2+x^2=12$
$\Rightarrow 4x^2=12$
$\Rightarrow x^2=3$
$\Rightarrow x=\pm \sqrt{3}$
So, coordinates of $B$ are $(\sqrt{3},1)$. Also, circle cut the $X$-axis at $A(2,0)$ and $Y$-axis at $(0,2)$
$[\because 2$ is radius of circle]
Required area $=$ Area of region $ODBO$
$+$ Area of region $DABD$
$=\underset{0}{\overset{\sqrt{3}}{\int }}y$ (line) $dx+\underset{\sqrt{3}}{\overset{2}{\int }}y$(circle) $dx$
$=\underset{0}{\overset{\sqrt{3}}{\int }}\frac{x}{\sqrt{3}}dx+\underset{\sqrt{3}}{\overset{2}{\int }}\sqrt{4-x^2}dx$
$=\frac{1}{\sqrt{3}}{\left[\frac{x^2}{2}\right]}_0^{\sqrt{3}}+\underset{\sqrt{3}}{\overset{2}{\int }}\sqrt{2^2-x^2}dx$
$=\frac{1}{\sqrt{3}}\left[\frac{{\left(\sqrt{3}\right)}^2-0}{2}\right]+{\left[\frac{x}{2}\sqrt{4-x^2}+\frac{4}{2}{\sin }^{-1}\left(\frac{x}{2}\right)\right]}_{\sqrt{3}}^2$
$=\frac{1}{\sqrt{3}}\times \frac{3}{2}+[\frac{2}{2}\sqrt{4-4}+2{\sin }^{-1}\left(\frac{2}{2}\right)$
$-\frac{\sqrt{3}}{2}\sqrt{4-3}-2{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)]$
$=\frac{\sqrt{3}}{2}+0+2{\sin }^{-1}\left(1\right)-\frac{\sqrt{3}}{2}-2{\sin }^{-1}\left(\frac{\sqrt{3}}{2}\right)$
$=2\times \frac{\pi }{2}-2\times \frac{\pi }{3}=\frac{\pi }{3}$