 Solution

Orient the body boundary counterclockwise and write its tangent displacement as $(dx,dy)$. The outward normal from the body has displacement-weighted components $(dy,-dx)$. Pressure therefore exerts the force components $F_x=-\oint pdy$ and $F_y=\oint pdx$, orFor steady irrotational incompressible flow at constant density, Bernoulli gives $p=p_0-\rho |w^{\prime }(z){|}^2/2$, with any constant far-field kinetic term absorbed into $p_0$. Its constant contribution integrates to zero. The boundary is a streamline, so $\mathrm{Im}w$ is constant there and $w^{\prime }(z)dz$ is real. Thus $w^{\prime }dz=\overline{w^{\prime }}d\bar{z}$, giving $|w^{\prime }{|}^{2}d\bar{z}=(w^{\prime }{)}^{2}dz$. This identity element remains valid at a stagnation point as both sides vanish. Substitution yields Blasius force theorem: