= Solution
Parametrize the path by $\mathbf r(t)=(x(t),y(t),z(t))$, from $t=a$ to $t=b$. For the <conservative vector field> $\mathbf F=\nabla\Phi$, the <chain rule> gives
$$
\mathbf F\cdot\frac{d\mathbf r}{dt}
=\Phi_x\frac{dx}{dt}+\Phi_y\frac{dy}{dt}+\Phi_z\frac{dz}{dt}
=\frac{d}{dt}\Phi(\mathbf r(t)).
$$
Integrating proves the <fundamental theorem for line integrals>:
$$
\boxed{\int_C\mathbf F\cdot d\mathbf r=\Phi(\mathbf r(b))-\Phi(\mathbf r(a))}.
$$
Only the endpoints matter, so the integral is independent of the connecting path.
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