Line integral independent of path

Statement. Let F be continuous in a region R. Then ∫_C F·dr between two points is independent of the path in R if and only if F = ∇φ for some single-valued scalar function φ with continuous first partial derivatives.

(⇐) If F = ∇φ, the integral is independent of path

Let C join A and B, with r(t) the position vector, t from t₁ to t₂. Then F·dr = ∇φ·dr = dφ, so

∫_C F·dr = ∫_C dφ = φ(B) − φ(A)

This depends only on the end points A and B, not on the path C.

(⇒) If the integral is independent of path, then F = ∇φ

Let F = F₁ i + F₂ j + F₃ k. Fix a point A in R and define for any point P(x, y, z) in R

φ(x, y, z) = ∫_A^P F·dr

By hypothesis this does not depend on the path, so φ is a single-valued function of P. Now compute ∂φ/∂x. Take a point P′(x + Δx, y, z) near P and choose the path from A to P′ as: A to P by any path, then the straight segment from P to P′ parallel to the x-axis. On that segment dy = dz = 0, so

φ(x + Δx, y, z) − φ(x, y, z) = ∫_P^{P′} F₁ dx = ∫_x^{x+Δx} F₁(s, y, z) ds

Divide by Δx and let Δx → 0. Since F₁ is continuous, the mean-value theorem gives

∂φ/∂x = F₁

In the same way, ∂φ/∂y = F₂ and ∂φ/∂z = F₃. Therefore

∇φ = ∂φ/∂x i + ∂φ/∂y j + ∂φ/∂z k = F₁ i + F₂ j + F₃ k = F

Hence F = ∇φ, with φ single-valued with continuous first partial derivatives. ∎

(Such an F is called conservative; then ∇×F = ∇×∇φ = 0 and ∮ F·dr = 0 around every closed curve in R.)