= Solution
On each of the two radial legs, $\mathbf F=\mathbf a\times\mathbf r$ is perpendicular to the tangent. Each contribution is zero, giving $\boxed{\int\mathbf F\cdot d\mathbf r=0}$. This differs from the straight-path answer, confirming path dependence.
For $\mathbf G$, use $\mathbf r=(s,0,0)$ with $s$ decreasing from one to zero, followed by $\mathbf r=(0,s,0)$ with $s$ increasing from zero to one. The two contributions are
$$
\int_1^0\frac{s\,ds}{(1+s^2)^{3/2}}=-1+\frac1{\sqrt2},\qquad \int_0^1\frac{s\,ds}{(1+s^2)^{3/2}}=1-\frac1{\sqrt2}.
$$
They cancel: $\boxed{\int\mathbf G\cdot d\mathbf r=0}$. This agreement is a consequence of the <conservative vector field> property already proved, rather than evidence sufficient by itself to establish it.
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