the cylinder of surface area dS is:dF = - p.n.dS1.26where n is a unit vector normal to the surface, and the minus sign indicates that this force acts inward.Next, we integrate this over the entire surface of the cylinder to find the net force:F = - ∫s p.n.dS1.27For the cylinder, (in polar coordinates)dS = Rdq.dz 1.28with z along the axis of the cylinder. The unit vector depends upon the angle q as n = cosq I+ sinq j1.29then the x and y axis components of the force are:Fx = - Lx R ∫ p(q) cos q dq. 1.30Which is the equation of the drag.Fy = - Lx.R.∫ p(q) sinq dq.1.31Which is the equation of the lift.Note that the integrals are from 0 to 2pThe value Lx is the length of the cylinder in the z-direction. If we now substitute our expression for the pressure distribution on the surface of the cylinder (eq. 1.25) into our expression for the drag and for the lift, we have:Fx = -Lx.R ∫ rU(1-4sinq).cosq dq1.32Fx = -.Lx.R.rU.∫ (1-4sinq).cosq.dqBut ∫ (1-4sinq).cosq.dq = ∫ cosq.dq - ∫ 4sinq.cosq.dq∫ (1-4sinq).cosq.dq = [sinq](02p) [4/3.sin^3](02p)∫ (1-4sinq).cosq.dq = 01.33so from 1.32, we have:Fx = 0So we are arrived at the paradox of DAlemberts: a body in an irrotational, non-viscous and incompressible flow produces no drag.The results found here are applicable to any shapes; not only cylinder.CONCLUSIONPhysically, the fact that the drag is zero is due to the symmetry of the pressure field about the cylinder. The fluid is pushing as hard on the upstream side of the cylinder as on the downstream side, so there are no unbalanced forces which would lead to drag. Whether or not this is paradoxical is a matter of opinion. If there were a drag on the cylinder how would the energy be dissipated in a nonviscous fluid ? to really understand drag, we need to go beyond the nonviscous approximation and treat real fluids.So the paradox of DAlembert comes from the fact that he neglected the frictio...
