I have trouble to figure it out how to derive the formulae that calculates viscous drag coeffient using vorticity. Any one can help me?

Viscous Drag D = integral of stress tensor T over the body. For incompressible flows, T = mu*(def(u)). (mu is viscosity assumed to be constant and u is velocity vector) So,

D = mu * integral of def(u)*dS -------(a)

def(u) is the deformation tensor defined by grad(u)+(grad(u))^t where t means transpose, and dS is the outward normal area vector on the body surface. Using the facts that (1)div(u)=0 on the surface (2)(w)x(dS)=[grad(u)-(grad(u))^t]*dS where w is the vorticity vector and "x" means a cross product (3)(grad(u))^t*dS=0 on the surface, you can show that

D = (a) = mu * integral of (w)x(dS)

which is the desired result.

I learned this a long time ago, but the professor said that this formula was impractical since measuring vorticity vector was difficult. Would you tell me more about what kind of problem you are dealing with? Why do you not compute velocities?