The area of the smaller region enclosed by the curves ${\mathrm{y}}^2=8\mathrm{x}+4\text{ and }{\mathrm{x}}^2+{\mathrm{y}}^2+4\sqrt{3}\mathrm{x}-4=0$ is equal to

$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+4\sqrt{3}\mathrm{x}-4=0\\ \Rightarrow {\mathrm{x}}^2+4\sqrt{3}\mathrm{x}+12+{\mathrm{y}}^2=16\\ \Rightarrow {\left(x+2\sqrt{3}\right)}^2+{\mathrm{y}}^2=16\Rightarrow {\left(x+2\sqrt{3}\right)}^2=16-y^2\\ \Rightarrow x+2\sqrt{3}=\sqrt{16-y^2}\\ {\mathrm{y}}^2=8\mathrm{x}+4\end{array}$

Point of intersections are (0, 2) & (0, -2)

Both are symmetric about x-axis

$\text{ Area }=2{\int }_0^2 \left(\sqrt{16-{\mathrm{y}}^2}-2\sqrt{3}\right)-\left(\frac{{\mathrm{y}}^2-4}{8}\right)\mathrm{dy}$

On solving $\text{ Area }=\frac{1}{3}[8\mathrm{\pi }+4-12\sqrt{3}]$