Find the surface area generated by rotating the lemniscate $r^2=6cos(2)$ about the polar axis.

a$.2\sqrt[\phantom{2}]{6}$

b$.12(2-\sqrt[\phantom{2}]{2})$

c$.\sqrt[\phantom{2}]{6}$

d$.\frac{\sqrt[\phantom{2}]{6}}{10}$

e$.\frac{}{6}$