 Solution

Differentiate the velocity potential in polar coordinates:At $r=a$ the normal velocity is zero. At large radius these components tend to $U\cos \theta ,-U\sin \theta$, the polar components of uniform flow in the positive $x$ direction. The circulation around any concentric circle is ${\int }_0^{2\pi }{u}_{\theta }rd\theta =\kappa$. Although the potential's circulation term is multivalued, its velocity is single-valued and irrotational away from the excluded axis.

For steady inviscid constant-density flow, Bernoulli equation gives $p+\rho |u{|}^2/2=p_{\infty }+\rho U^2/2$. On the cylinder, $u_{\theta }=-2U\sin \theta +\kappa /(2\pi a)$, soIn the displayed force integral, every constant or ${\sin }^2\theta$ term has zero cosine and sine first moments. The cross term in pressure is $\rho U\kappa \sin \theta /(\pi a)$. Therefore, using exactly the positive-sign convention printed,per unit axial length, since ${\int }_0^{2\pi }{\sin }^2\theta d\theta =\pi$.

With the usual normal directed outward from the cylinder into the fluid, pressure exerts traction $-pn$, so the physical force of the fluid on the cylinder is instead $\boxed{(0,-\rho U\kappa )}$. The positive integral printed in the question has the opposite sign. This makes the convention explicit and is consistent with the Kutta–Joukowski theorem for positive counterclockwise circulation.