= Solution
Parametrize the segment by $\mathbf r(t)=(1-t,t,0)$, $0\leq t\leq1$. Its tangent is $(-1,1,0)$ and $\mathbf F=(-t,1-t,t-1)$, so $\mathbf F\cdot\mathbf r'=1$. Hence its <line integral> is
$$
\boxed{\int\mathbf F\cdot d\mathbf r=1.}
$$
For the other <line integral>, $|\mathbf r|^2=(1-t)^2+t^2$ and $\mathbf r\cdot\mathbf r'=2t-1$. Thus
$$
\int\mathbf G\cdot d\mathbf r=\int_0^1\frac{2t-1}{[1+(1-t)^2+t^2]^{3/2}}dt=\left[-\frac1{\sqrt{1+(1-t)^2+t^2}}\right]_0^1=\boxed0.
$$
This also equals the change in $\Phi$, by the <fundamental theorem for line integrals>.
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