Dedicated to Comrade Otto Bretscher.

This is an example of the Divergence Theorem. If we consider the closed surface $T$ consisting of $S$ together with the unit square defined by $0 \le x,y \le 1$ and $z=0$ , then $\oint_T x\mathbf{k} \cdot d\mathbf{A} \; = \; \iiint_V \nabla \cdot x\mathbf{k}\, dV \; = \; \iiint_V(0+0+\tfrac{\partial x}{\partial z}) \,dV \; = \; 0$ where $V$ is the volume enclosed by $T$ . But the outward-pointing normal for the unit square is $-\mathbf{k}$ and so $\oint_T x\mathbf{k} \cdot d\mathbf{A} \; = \; \iint_S x\mathbf{k} \cdot d\mathbf{A} - \int_0^1 \int_0^1 x\,dx\,dy \; = \; \iint_S x\mathbf{k} \cdot d\mathbf{A} - \tfrac12 \;,$ giving us the answer.