= Solution
With polar unit vectors $\mathbf e_r=(\cos\theta,\sin\theta,0)$ and $\mathbf e_\theta=(-\sin\theta,\cos\theta,0)$, the second term is $\Gamma\mathbf e_\theta/(2\pi r)$. It is tangent to circles, independent of $z$, and has zero <divergence> and zero <vorticity> away from the axis. Its singular axis carries a <line vortex>; adding the constant <velocity field> gives a <uniform flow> plus that <line vortex>.
The counterclockwise <circulation> around a positively oriented loop is $\oint\mathbf u\cdot d\mathbf r$. For a radius-$R$ circle, the <uniform flow> contributes zero and the <line vortex> contributes
$$
\oint_{C_R}\frac\Gamma{2\pi R}\mathbf e_\theta\cdot R\mathbf e_\theta\,d\theta=\Gamma.
$$
Any loop winding once around the axis has the same <circulation>, by <Stokes theorem> in the nonsingular intervening region.
On $C_R$, $\mathbf n=\mathbf e_r$, $\mathbf u\cdot\mathbf n=U\cos\theta$, and $dl=R\,d\theta$. Thus
$$
\oint_{C_R}(\mathbf u\cdot\mathbf n)\mathbf u\,dl=RU^2\mathbf e_x\int_0^{2\pi}\cos\theta\,d\theta+\frac{U\Gamma}{2\pi}\int_0^{2\pi}\cos\theta\,\mathbf e_\theta\,d\theta=\boxed{\frac12\boldsymbol\Gamma\mathbin\times\mathbf U}.
$$
In fact this equality holds for every $R>0$ for the stated field, since the remaining vector <integral> is $(0,\pi,0)$.
This is the advective part of the <momentum flux>. To obtain the force, include <pressure>. For steady <irrotational flow>, the <Bernoulli equation> gives
$$
p=p_\infty-\frac\rho2(|\mathbf u|^2-U^2)=p_\infty+\frac{\rho U\Gamma\sin\theta}{2\pi R}-\frac{\rho\Gamma^2}{8\pi^2R^2}.
$$
The constant terms integrate to zero against $\mathbf n$; the remaining <pressure> contribution is $\oint p\mathbf n\,dl=\rho\boldsymbol\Gamma\times\mathbf U/2$. The total outward <momentum flux> plus <pressure> term is therefore $\rho\boldsymbol\Gamma\times\mathbf U$, the force of the obstacle on the fluid. The opposite force, of fluid on the obstacle per unit span, is
$$
\boxed{\mathbf F=\rho\mathbf U\times\boldsymbol\Gamma.}
$$
This is the <Kutta–Joukowski theorem>, with positive $\Gamma$ defined counterclockwise: positive $U$ and $\Gamma$ give force in the negative $y$ direction. For an actual two-dimensional <aerofoil>, faster-decaying far-field terms do not change this limit. \b[The displayed advective integral supplies half the lift; pressure supplies the other half.]
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