The area of the smaller region enclosed by the curves y ^ 2 = 8x +...

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 $= 2{∫}_{0}^{2}\left(\sqrt{16 - {y}^{2}} - 2\sqrt{3}\right) - \left(\dfrac{{y}^{2} - 4}{8}\right)dy$

$= 2{\left[\dfrac{1}{2}\left(y\sqrt{16 - {y}^{2}} + 16\sin^{- 1}\left(\dfrac{y}{4}\right)\right) - 2\sqrt{3}y - \dfrac{{y}^{3}}{24} + \dfrac{1}{2}y\right]}_{0}^{2}$

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