---
title: "State and prove Stokes' theorem"
subject: "Electromagnetism"
chapter: 1
key: "semester_3/electromagnetism:stokes-theorem-state-prove"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/stokes-theorem-state-prove
site: BCT Engineering
---

# State and prove Stokes' theorem

Electromagnetism · Chapter 1: Introduction

- Asked in: 2076 Chaitra, 5 marks; 2070 Chaitra, 2 marks; 2068 Baisakh, 3 marks (asked 3 times)
- Video: [Stokes Theorem : Basics, Proof, Physical Significance and Applications](https://www.youtube.com/watch?v=D2eHgZ4kMHU) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/stokes-theorem-state-prove
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-1

## Question

State and prove Stokes' theorem.

## Short answer

Stokes' theorem equates the line integral of a vector field around a closed loop to the surface integral of its curl: ∮A·dl = ∫(∇×A)·dS. Proof partitions the surface into small loops; interior edges cancel, leaving the boundary, and the sum of circulations becomes the surface integral of curl.

## Answer

**Statement:** The line integral of a vector field **A** around a closed path L equals the surface integral of the curl of **A** over any open surface S bounded by L.

```text
∮_L A · dl = ∫_S (∇ × A) · dS
```

The direction of dS (normal to S) is related to the direction of travel along L by the right-hand rule.

#### Proof

```text
        ┌────┬────┬────┐   S is divided into many
        │ ΔS │ ΔS │ ΔS │   small patches ΔS_k.
        ├────┼────┼────┤   Each patch has boundary
        │ ΔS │ ΔS │ ΔS │   ΔL_k, traversed in the
        └────┴────┴────┘   same sense as L.
```

1. Divide the open surface S into a very large number of small patches, each of area ΔS_k, bounded by a small closed path ΔL_k, all traversed in the same (counter-clockwise) sense.
2. For one small rectangular patch Δx × Δy in the xy-plane, the circulation of **A** around it (the four sides) is
   ```text
   ∮ A · dl ≈ A_x(y)Δx + A_y(x+Δx)Δy − A_x(y+Δy)Δx − A_y(x)Δy
            = (∂A_y/∂x − ∂A_x/∂y) Δx Δy
            = (∇ × A)_z ΔS
   ```
   So for any small patch: ∮ A · dl = (∇ × A) · ΔS_k.
3. Add this over all patches. On every interior common edge, the two neighbouring patches traverse the edge in opposite directions, so the contributions cancel. Only the edges lying on the outer boundary L survive:
   ```text
   Σ_k ∮_ΔL_k A · dl = ∮_L A · dl
   ```
4. In the limit ΔS_k → 0 the sum on the right-hand side becomes an integral:
   ```text
   ∮_L A · dl = Σ_k (∇ × A) · ΔS_k  →  ∫_S (∇ × A) · dS
   ```

Hence ∮ A · dl = ∫ (∇ × A) · dS. **Proved.**

**Note:** The theorem converts a line integral into a surface integral (and vice versa). It holds for any surface bounded by L, so the result does not depend on the shape of S. It is used to derive Maxwell's equation ∮ H · dl = I from ∇ × H = J.

## Check yourself

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

1. What does Stokes' theorem equate?
   A. The closed line integral of A around L equals the surface integral of curl A over any surface bounded by L
   B. The closed surface integral of A equals the volume integral of curl A
   C. The closed line integral of A equals the surface integral of div A
   D. The surface integral of A equals the line integral of curl A
   Answer: A. Stokes' theorem converts the circulation of A around a closed path into the flux of curl A through any open surface bounded by that path. (Key point: Statement of Stokes' theorem)

2. In the proof, for a small rectangle in the xy-plane the circulation of A around it works out to what?
   A. (dA_x/dx + dA_y/dy) dx dy
   B. (dA_y/dx - dA_x/dy) dx dy = (curl A)_z dS
   C. A_x dx + A_y dy
   D. (dA_z/dx - dA_x/dz) dx dy
   Answer: B. Adding the four sides gives (dA_y/dx - dA_x/dy) dx dy, which is the z-component of curl A times the patch area. (Key point: Circulation of a small patch)

3. Why does summing over all patches leave only the line integral around the outer boundary L?
   A. Interior edges have zero field
   B. Interior patches have zero curl
   C. Each interior edge is shared by two patches traversed in opposite directions, so contributions cancel
   D. The patches are made so small that interior edges vanish
   Answer: C. A common edge is traversed in opposite senses by neighbouring patches, so only the outer boundary edges survive. (Key point: Cancellation of interior edges)

4. What does the theorem tell us about the choice of surface S bounded by L?
   A. S must be flat
   B. The result depends on the area of S
   C. S must be a closed surface
   D. Any open surface bounded by L gives the same value, so the shape of S does not matter
   Answer: D. The left side depends only on the boundary L, so every surface bounded by L gives the same flux of curl A. (Key point: Independence of surface shape)

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