Line integral independent of path

Part 1: F = ∇φ implies path independence.

Let A and B be two points in the region, joined by any smooth curve C given by r = r(t), t from t₁ to t₂. Since F = ∇φ = (∂φ/∂x) i + (∂φ/∂y) j + (∂φ/∂z) k,

∫_C F·dr = ∫_C (∂φ/∂x dx + ∂φ/∂y dy + ∂φ/∂z dz)
         = ∫_C dφ
         = φ(B) − φ(A)

The result depends only on the end points A and B, not on the curve joining them. Hence the line integral is independent of path (and for a closed curve, ∮ F·dr = 0).

Part 2: Converse. Suppose ∫_A^B F·dr is independent of the path in the region, and write F = P i + Q j + R k. Fix a point A and define

φ(x,y,z) = ∫_A^(x,y,z) F·dr  (any path)

This is well defined by path independence. To find ∂φ/∂x, go from (x,y,z) to (x + Δx, y, z) along a straight segment parallel to the x-axis, where dy = dz = 0:

φ(x+Δx,y,z) − φ(x,y,z) = ∫_x^(x+Δx) P(s,y,z) ds

Divide by Δx and let Δx → 0. Since P is continuous, the right side tends to P(x, y, z), so ∂φ/∂x = P. In the same way ∂φ/∂y = Q and ∂φ/∂z = R. Therefore

F = P i + Q j + R k = ∇φ,

so F is the gradient of a scalar function. Also curl F = curl(∇φ) = 0.

Thus ∫ F·dr is path independent if and only if F = ∇φ.