= 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> is
$$
\partial_xF_y-\partial_yF_x=\frac{y^2-x^2}{(x^2+y^2)^2}-\frac{y^2-x^2}{(x^2+y^2)^2}=0.
$$
Thus \b[$\nabla\times\mathbf F=0$ 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 $2\sqrt2>1$. The field is continuously differentiable on a neighborhood of that disk. The <Stokes theorem> therefore gives
$$
\boxed{\oint_{\gamma_1}\mathbf F\cdot d\mathbf x=0.}
$$
This value is independent of the choice of orientation.
For $\gamma_2$, take the counterclockwise orientation viewed from positive $z$. Put $\mathbf x(t)=(\cos t,\sin t,0)$, $0\leq t\leq2\pi$. Then $\mathbf F(\mathbf x(t))=(-\sin t,\cos t,0)=\mathbf x'(t)$, so the <line integral> is
$$
\boxed{\oint_{\gamma_2}\mathbf F\cdot d\mathbf x=\int_0^{2\pi}1\,dt=2\pi.}
$$
Clockwise 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.
Back to article page
