 Solution

The product rule, the Laplace equation and $\mathbf{u}=\nabla \varphi$ giveFirst truncate the exterior fluid domain by a large circle $C_R$. Let ${\mathbf{n}}_B$ point outward from the body into the fluid. The outward normal of the exterior fluid domain at the body is $-{\mathbf{n}}_B$. The planar divergence theorem, equivalently Green's theorem, therefore givesOrient $\partial B$ counterclockwise around the body. Then ${\mathbf{n}}_{B}ds=(dy,-dx)$ andFor the finite-energy exterior rigid-body flow intended here, the outer boundary term vanishes as $R\to \infty$. In particular, zero net volume flux and zero circulation eliminate the $1/z$ term in the exterior Laurent series of the complex velocity. Thus $\mathbf{u}=O(R^{-2})$ and, after choosing the additive potential constant, $\varphi =O(R^{-1})$. The outer integral is then $O(R^{-2})$.

Multiplying by half the constant mass density gives the boundary integral for kinetic energy of planar potential flow:The minus sign comes from the fluid's inner-boundary orientation, rather than a negative physical kinetic energy.

There is a necessary qualification to the printed hypotheses. Decaying velocity and zero circulation alone do not ensure a finite kinetic energy: $\varphi =A\log r$ gives $\mathbf{u}=A{\mathbf{e}}_r/r$, zero circulation, and logarithmically divergent exterior kinetic energy. It carries a nonzero volume flux from the body and is not the flow of an impermeable rigid body. The finite-energy condition for planar exterior potential flow makes explicit the source-free assumption needed to discard the outer boundary term.