= Solution
{c}
For an oriented piecewise smooth <parametric curve> $\mathbf r:[a,b]\to\mathbb R^3$ and a continuous <vector field> $\mathbf F$ along it, the vector-field <line integral> is
$$
\boxed{\int_C\mathbf F\cdot d\mathbf r=\int_a^b\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)\,dt.}
$$
For a piecewise smooth curve, sum these integrals over its smooth pieces. The <dot product> selects the component of the <vector field> in the direction of motion.
If $\mathbf F=\nabla\phi$ for a <smooth function> $\phi$, the <multivariable chain rule> gives
$$
\mathbf F(\mathbf r(t))\cdot\mathbf r'(t)=\frac{d}{dt}\phi(\mathbf r(t)).
$$
The <fundamental theorem of calculus>, with cancellation at intermediate endpoints, therefore proves the <fundamental theorem for line integrals>:
$$
\int_C\nabla\phi\cdot d\mathbf r=\phi(\mathbf r(b))-\phi(\mathbf r(a)).
$$
On a closed curve $\mathbf r(a)=\mathbf r(b)$, so $\boxed{\oint_C\nabla\phi\cdot d\mathbf r=0}$.
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