= Solution
The <position vector> field is a <gradient field>:
$$
\mathbf r=(x,y,z)=\nabla\left(\frac{x^2+y^2+z^2}{2}\right).
$$
Along any piecewise smooth <curve>, $\mathbf r\cdot d\mathbf r=d(|\mathbf r|^2/2)$. Therefore a closed <line integral> has identical endpoint values and
$$
\boxed{\oint_C\mathbf r\cdot d\mathbf r=\left[\frac{|\mathbf r|^2}{2}\right]_{\mathrm{start}}^{\mathrm{end}}=0.}
$$
For a loop bounding an admissible <oriented surface>, <Stokes theorem> gives the same result because $\nabla\times\mathbf r=\mathbf0$. The <gradient field> proof also applies directly to arbitrary piecewise smooth closed loops without choosing a spanning surface.
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