The area of the entire sector is [tex](\pi)(6^{2}) \left(\frac{60}{360} \right)=6\pi[/tex]
The area of the triangle OAB is [tex]\frac{\sqrt{3}}{4}(6^{2})=9\sqrt{3}[/tex].
So, the area of the segment is [tex]\boxed{6\pi-9\sqrt{3}}[/tex]
