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:
\( \dfrac{1}{3}\left(2-12\sqrt{3}+8\pi\right) \)
\( \dfrac{1}{3}\left(2-12\sqrt{3}+6\pi\right) \)
\( \dfrac{1}{3}\left(4-12\sqrt{3}+8\pi\right) \)
\( \dfrac{1}{3}\left(4-12\sqrt{3}+6\pi\right) \)