 Solution

Parametrize the segment by $\mathbf{r}(t)=(1-t,t,0)$, $0\le t\le 1$, so ${\mathbf{r}}^{\prime }=(-1,1,0)$. For the first field, ${\mathbf{F}}_1=(-t,1-t,-1+t)$ and its dot product with the tangent is one. ThereforeFor the second field, use the fundamental theorem for line integrals. The endpoints both have $r=1$, so their potential values agree andDirectly, its parameter integrand is $(2t-1)/[1+(1-t{)}^2+t^2{]}^{3/2}$, which changes sign under $t\mapsto 1-t$ and integrates to zero.