Question

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

Answer 1

Given equation of circle is $x^2+y^2=4$ and $X=\sqrt{3}yory=\frac{1}{\sqrt{3}}x$ represents a line through the origin.

The line $y=\frac{1}{\sqrt{3}}x$ intersect the circle so it will satisfy the equation of circle

$\therefore x^2+(\frac{1}{\sqrt{3}}x{)}^2=4\Rightarrow \frac{4}{3}{x}^2=\frac{4\times 3}{4}=3\Rightarrow x=\pm \sqrt{3}$

When $x=\sqrt{3},$ then $y=\frac{1}{\sqrt{3}}\left(\sqrt{3}\right)=1$

When $x=\sqrt{3}$, then $y=\frac{1}{\sqrt{3}}\left(\sqrt{3}\right)=1.$

[For first quadrant we take $x=\sqrt{3}$ and neglect \(x = - \sqrt 3.]

$\therefore$ The line and the circle meet at the point $(\sqrt{3},1).$

Required area (shaded region in first quadrant)

= (Area under the line $y=\frac{1}{\sqrt{3}}x$ from x = 0 to $x=\sqrt{3})$

+ (Area under the circle from $x=\sqrt{3}tox=2)$

$={\int }_0^{\sqrt{3}}\frac{1}{\sqrt{3}}xdx+{\int }_{\sqrt{3}}^2\sqrt{4+-x^2}dx(\because x^2+y^2=4\Rightarrow y=\sqrt{4-x^2})=\frac{1}{\sqrt{3}}.[\frac{x^2}{2}{]}_0^{\sqrt{3}}+[\frac{x}{2}\sqrt{4-x^2}+\frac{2^2}{2}si{n}^{-1}\left(\frac{x}{2}\right){]}_{\sqrt{3}}^2=\frac{1}{2\sqrt{3}}\left[\right(\sqrt{3}{)}^2-0^2]+[0+2si{n}^{-1}\left(1\right)-\frac{\sqrt{3}}{2}\sqrt{4-3}-2si{n}^{-1}\left(\frac{\sqrt{3}}{2}\right)]=\frac{\sqrt{3}}{2}+2(\frac{\pi }{2})-\frac{\sqrt{3}}{2}-2(\frac{\pi }{3})=\pi -\frac{2\pi }{3}=\frac{\pi }{3}squnit$