= Solution
The shorter circular arc lies in the $x$-$y$ <plane> and can be written $\mathbf r(\theta)=(\cos\theta,\sin\theta,0)$, $0\leq\theta\leq\pi/2$. The first field is $(-\sin\theta,\cos\theta,-\cos\theta)$, whose <dot product> with $\mathbf r'$ is one. Hence
$$
\boxed{\int_{\rm arc}\mathbf F_1\cdot d\mathbf r=\frac\pi2}.
$$
The second field is radial while the circular tangent is perpendicular to the radius, so $\boxed{\int_{\rm arc}\mathbf F_2\cdot d\mathbf r=0}$. Thus the first field gives different values, $1$ and $\pi/2$, along the two paths, whereas the conservative second field gives the same endpoint difference.
As a check using <Stokes theorem>, the arc followed by the reversed segment encloses <area> $\pi/4-1/2$ in the <plane>. The constant <curl> has upward component two, giving circulation $2(\pi/4-1/2)=\pi/2-1$, exactly the difference above.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2010/paper-2-line-integral-paths.png]
{title=Straight and circular paths with the same endpoints, enclosing the area responsible for their circulation difference}
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