Based on Stewart - Calculus
Is this proof correct?
If $F$ is conservative then $\exists f$ s.t. $F = \nabla f$. If $C$ is closed, then r(b)=r(a) hence RHS of the fundamental theorem of line integrals is zero. QED
2026-03-27 05:36:31.1774589791
Prove conservative implies zero line integral over closed without path independence
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$$ \int_{\zeta }f(r)dr = \int_{\zeta} \nabla \varphi dr = \int_{\zeta} d \varphi = \varphi_B -\varphi_A \to \oint f(r) dr = \varphi_A - \varphi_A = 0 $$