 Solution

Write $r^2=x^2+y^2$. The first two components have no $z$ dependence, and the third component is zero. The only possibly nonzero component of the curl isThus $\nabla \times \mathbf{F}=0$ wherever the field is defined.

The disk spanning ${\gamma }_1$ lies away from the excluded axis: the distance from its center to that axis is $2\sqrt{2}>1$. The field is continuously differentiable on a neighborhood of that disk. The Stokes theorem therefore givesThis value is independent of the choice of orientation.

For ${\gamma }_2$, take the counterclockwise orientation viewed from positive $z$. Put $\mathbf{x}(t)=(\cos t,\sin t,0)$, $0\le t\le 2\pi$. Then $\mathbf{F}(\mathbf{x}(t))=(-\sin t,\cos t,0)={\mathbf{x}}^{\prime }(t)$, so the line integral isClockwise orientation gives $-2\pi$; no orientation is printed for this curve. There is no contradiction with the Stokes theorem: its usual spanning disk intersects the axis where the field is undefined, so the differentiability hypothesis fails. This is a curl-free vector field with nonzero circulation around the excluded axis; local angular potentials cannot be combined into a single-valued global potential.