Question

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

Answer 1

x-axis, $x=\sqrt{3}y$,

$x^2+y^2=4$

$x^2+y^2=r^2$

$\Rightarrow r=2$

Given $r=2$ & $y=\frac{1}{\sqrt{3}}x$

$\Rightarrow y=mx+c$

$\therefore m=\frac{1}{\sqrt{3}}=\tan \theta$

$\theta ={30}^o$

So, area of shaded region

$\Rightarrow \frac{30}{360}\times \pi (2{)}^2$

$\Rightarrow \frac{1}{12}\times 4\pi$

$\Rightarrow \frac{\pi }{3}$.