Drag force and Stokes' force The drag force equation is,
$$F=1/2pAV^2$$
which means that it's proportional to the square of velocity. For Stokes' law, the drag force equation is,
$$F=6\pi\mu RV$$
which is proportional to the velocity (not it's square). What is wrong?
 A: The drag equation is an expression for bodies moving at a high-enough Reynolds number than the flow behind them is turbulent. It comes from dimensional analysis where it is assumed that there is a non-dimensional constant ($c_d$) that depends on the drag force, the shape of the body, the fluid the body is moving in, and how fast it moves. All of those parameters may only be combined in one way and have a non-dimensional result. So you start from defining a non-dimensional number, and then rearrange.
Stokes law, on the other hand, comes from integrating the pressure and viscous forces on a small body in very low speed flow. It is exact -- that is what the equations of motion give you. And because it is valid only at very low Reynolds numbers, there is no reason to expect it to have the same form as the drag equation for high Reynolds number flow. 
The other thing to consider is that the drag coefficient is not constant and does depend on flow conditions and velocity. So you can equate the two expressions, substitute in your area as a function of radius (the same radius used in Stokes law) and then compute an equivalent $c_d$ to use in the drag equation. You will get something of the form $c_d \propto 1/Re$.
