Strict general mathematical definition of drag Is there a formal definition of drag, say, as some surface integral of normal and shear forces? There seem to be a lot of formulas for specific cases, but is there a general one?
I need to accurately calculate the drag of three cylinders placed between two plates. 
 A: The drag of a fluid acting on an object inside is the flow of momentum through the boundary of the object. The momentum conservation law is the entire content of the Navier stokes equation, which can be written in integral form:
$$ {\partial\over \partial t} \int_R \rho v^i  = - \int_{\partial R} \rho v^i v\cdot \hat{n} + \int_{\partial R} (P \hat{n} + \nu(\rho) \nabla v^i )\cdot \hat{n} $$
Where $\hat{n}$ is the normal to the boundary of $R$, $P$ is the pressure, $\nu$ is the viscosity (as a function of the density $\rho$), and v is the velocity.
The left hand side says that you are looking at the flow of total i-component  of momentum out of region R. The first term on the right is the physical amount of momentum flowing out of the boundary of R by the flow of the fluid. The last term is the flow of momentum through the boundary of R due to forces at the edge.
Using the divergence theorem, you learn that
$$ \int_R {\partial\over\partial t} (\rho v^i) + \partial_j(\rho v^i v^j) - \partial_i P - \nabla\cdot(\nu \nabla v^i ) d^dx = 0$$
And you conclude that the NS equations are satisfied.
$$ {\partial\over\partial t} (\rho v^i) + \partial_j(\rho v^i v^j) - \partial_i P -\nabla \cdot (\nu \nabla v^i) $$
If you expand this out, and use the continuity equation, you will recover the more standard forms, but this is the form in which it is most transparently a continuity equation for the momentum flow.
So you see that the flow of the i-component of momentum into any region R due to the fluid, which is the i-th component of the force exerted by the fluid on whatever is inside R, is given by the boundary integral
$$ F^i_R = - \int_{\partial R} \rho v^i v\cdot \hat{n} + \int_{\partial R} (P\hat{n} + \nu \nabla v^i) \cdot \hat{n}$$
For the case where you have a solid object that the fluid cannot penetrate, the velocity is perpendicular to the object's surface, and the first term is zero (obviously-- the first term describes the momentum carried along with the fluid, and this is not entering R).
So the drag is the integral of two terms across the surface, the pressure across the object, which tells you how much the object is pushing to get the water to go around, and the gradient of the velocity, which describes how the viscosity pulls the object. For a moving object, this works at one instant to tell you how much momentum is entering or leaving the object, which is the instantaneous drag force.
A: Fist consider laminar flow and then you consider the boundary layer of fluid near the solids. For simple cases you use the incompressible Navier-Stokes to find the force balance which will yield your drag force. Typically this is done through CFD (Computational Fluid Dynamics) software.
I doubt you can solve your problem by hand as it is only feasible for the simplest of cases where the flow lines are parallel to the geometry and integration is possible.
