---
title: "Line integral independent of path"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:line-integral-independent-of-path-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/line-integral-independent-of-path-two
site: BCT Engineering
---

# Line integral independent of path

Mathematics III · Chapter 2: Line, Surface and Volume Integrals

- Asked in: 2079 Bhadra, 5 marks; 2072 Chaitra, 5 marks (asked 2 times)
- Video: [Path independence for line integrals | Multivariable Calculus | Khan Academy](https://www.youtube.com/watch?v=K_fgnCJOI8I) — Khan Academy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/line-integral-independent-of-path-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Prove that the line integral ∫_C F·dr of a continuous vector function F defined in a region R is independent of the path C joining any two points in R if and only if there exists a single valued scalar function φ, having first order partial derivatives, such that F = ∇φ.

## Short answer

If F = ∇φ, then ∫_C F·dr = φ(B) − φ(A), independent of path. Conversely, if the integral is independent of path, define φ(P) = ∫_A^P F·dr; then ∂φ/∂x = F₁, etc., so F = ∇φ.

## Answer

**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

```text
∫_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

```text
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

```text
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

```text
∂φ/∂x = F₁
```

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

```text
∇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.)

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. If F = grad phi, why is the integral path independent?
   A. F.dr = d(phi), so the integral is phi(B) - phi(A)
   B. Because F is constant
   C. Because curl F is nonzero
   D. Because dr = 0
   Answer: A. The integral depends only on end values of phi. (Key point: Easy direction)

2. How is phi defined in the converse?
   A. phi = |F|
   B. phi(P) = integral of F.dr from fixed point A to P
   C. phi = div F
   D. phi = F.r at P only
   Answer: B. Path independence makes this a well-defined single-valued function. (Key point: Converse definition)

3. How is dphi/dx found?
   A. Differentiate F1 w.r.t. x
   B. Move along a diagonal
   C. Move P to P' parallel to the x-axis (dy = dz = 0) and take the limit of the difference quotient
   D. Take the curl of F
   Answer: C. The difference reduces to the integral of F1 dx, whose limit is F1 by continuity. (Key point: Partial derivative)

4. What does the argument give?
   A. F = curl phi
   B. F = div phi
   C. phi = 0
   D. dphi/dx = F1, dphi/dy = F2, dphi/dz = F3, so F = grad phi
   Answer: D. All three components match, giving F = grad phi. (Key point: Conclusion)

## Related in Mathematics III

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