The area of the smaller region enclosed by the curves y 2 = 8 x + 4 and x 2 + y 2 + 4 3 x - 4 = 0 is equal to

Plotting the area of the smaller region enclosed by the curves $y^2=8x+4$ and $x^2+y^2+4\sqrt{3}x-4=0$ we get,

Now finding point of intersection of $x^2+y^2+4\sqrt{3}x-4=0$ and 

$y^2=8x+4$

We get, $\left(0,2\right)$ and $\left(0,-2\right)$

Both are symmetric about $x-$axis

So, area will be $=2{\int }_0^2\left(\sqrt{16-y^2}-2\sqrt{3}\right)-\left(\frac{y^2-4}{8}\right)dy$

$=2{\left[\frac{1}{2}\left(y\sqrt{16-y^2}+16{\sin }^{-1}\left(\frac{y}{4}\right)\right)-2\sqrt{3}y-\frac{y^3}{24}+\frac{1}{2}y\right]}_0^2$

$=\frac{1}{3}\left[8\pi +4-12\sqrt{3}\right]$