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

A

$$\frac{1}{3}\left(2 - 12\sqrt{3} + 8\pi\right)$$

B

$$\frac{1}{3}\left(2 - 12\sqrt{3} + 6\pi\right)$$

C

$$\frac{1}{3}\left(4 - 12\sqrt{3} + 8\pi\right)$$

D

$$\frac{1}{3}\left(4 - 12\sqrt{3} + 6\pi\right)$$