 Solution

Choose counterclockwise traversal viewed from $+z$, and span the curve by the disk $z=1$, $x^2+y^2\le 2$ with upward normal. HereThe surface side is therefore $-{\iint }_D{x}^{2}dxdy$. In polar coordinates it isFor the direct boundary integral, $dz=0$ gives $\nabla \psi \cdot d\mathbf{l}=dx$ on $z=1$. Set $x=\sqrt{2}\cos \theta$, $y=\sqrt{2}\sin \theta$, soThe two independent evaluations agree. Clockwise traversal and downward normal give $+\pi$ on both sides.