The area (in sq. units) of the smaller portion enclosed between the curves, x 2 + y 2 = 4 and y 2 = 3 x , is:

To find point of intersection, eliminate $y$ from both the equation, we get

$x^2+3x-4=0$

$\Rightarrow \left(x+4\right) \left(x-1\right)=0$

$x=-4, x=1$

Area $=\left(\underset{0}{\overset{1}{\int }}\sqrt{3}·\sqrt{x}·dx+\underset{1}{\overset{2}{\int }}\sqrt{4-x^2}·dx\right)\times 2$

$=\left(\sqrt{3}{\left(\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right)}_0^1+{\left(\frac{x}{2}\sqrt{4-x^2}+2{\sin }^{-1}\frac{x}{2}\right)}_1^2\right)\times 2$

$=\left(\sqrt{3}\left(\frac{2}{3}\right)+\left\{2·\frac{\pi }{2}-\left(\frac{\sqrt{3}}{2}+\frac{\pi }{3}\right)\right\}\right)\times 2$

$=\left(\frac{2}{\sqrt{3}}-\frac{\sqrt{3}}{2}+\frac{2\pi }{3}\right)\times 2$

$=\left(\frac{1}{2\sqrt{3}}+\frac{2\pi }{3}\right)\times 2=\frac{1}{\sqrt{3}}+\frac{4\pi }{3}$ sq. units