---
title: "Evaluate Stokes' theorem for G field"
subject: "Electromagnetism"
chapter: 1
key: "semester_3/electromagnetism:stokes-theorem-evaluate-g-field"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/stokes-theorem-evaluate-g-field
site: BCT Engineering
---

# Evaluate Stokes' theorem for G field

Electromagnetism · Chapter 1: Introduction

- Asked in: 2075 Asoj, 1+7 marks
- 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-evaluate-g-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-1

## Question

State Stokes' theorem. Evaluate both sides of Stokes' theorem for the field G = 10 sinθ a_φ and the surface r = 3, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ 90°. Let the surface have the a_r direction.

## Short answer

∮G·dl = ∫(∇×G)·dS. For G=10 sinθ a_φ on r=3, 0≤θ≤90°, 0≤φ≤90°: left side integrates over equator arc only, giving 30×(π/2)=15π. Right side: (∇×G)_r = 20 cosθ/r, flux = 15π. Both sides equal 15π ≈ 47.1.

## Answer

**Stokes' theorem:** The circulation of a vector field around a closed path equals the flux of its curl through any surface bounded by that path: ∮ G · dl = ∫ (∇ × G) · dS.

**Assumption:** A polar angle θ cannot exceed π, so the range is taken as 0 ≤ θ ≤ 90° (the usual form of this problem). The surface is the spherical patch r = 3, 0 ≤ θ ≤ 90°, 0 ≤ φ ≤ 90° (one octant of a sphere), with dS along a_r.

```text
          z
          │   patch: r = 3
          │ ╭──╮
          │╱    ╲  boundary = equator arc (θ = 90°, φ: 0→90°)
          ●──────── y         + two meridian arcs (φ = 0 and φ = 90°)
         ╱
        x
```

#### Left side: closed line integral

G = 10 sinθ a_φ has only an a_φ component.
- On the two meridian arcs (φ = 0, φ = 90°), dl is along a_θ, so G · dl = 0.
- On the equator arc (r = 3, θ = 90°), G_φ = 10 sin90° = 10 and dl = r sinθ dφ a_φ = 3 dφ a_φ.

```text
∮ G · dl = ∫₀^(π/2) 10 (3) dφ = 30 (π/2) = 15π = 47.12
```

#### Right side: surface integral of curl

```text
(∇ × G)_r = (1/(r sinθ)) ∂(sinθ G_φ)/∂θ = (1/(r sinθ)) (10 · 2 sinθ cosθ) = 20 cosθ / r
(∇ × G)_θ = −(1/r) ∂(r G_φ)/∂r = −10 sinθ / r
```

With dS = r² sinθ dθ dφ a_r only the a_r component contributes:

```text
∫ (∇ × G) · dS = ∫₀^(π/2) ∫₀^(π/2) (20 cosθ / 3) (9 sinθ) dθ dφ
               = 60 ∫₀^(π/2) sinθ cosθ dθ · ∫₀^(π/2) dφ
               = 60 (1/2)(π/2) = 15π = 47.12
```

**Answer:** Both sides equal **15π ≈ 47.1** (in the units of G). Stokes' theorem is verified.

## Check yourself

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

1. Which part of the boundary contributes to the line integral of G = 10 sinθ a_φ?
   A. The two meridian arcs only
   B. Only the equator arc at θ = 90°, where dl is along a_φ
   C. All three arcs equally
   D. None, since G is zero on the boundary
   Answer: B. On the meridian arcs dl is along a_θ so G·dl = 0, leaving the equator arc where G_φ = 10 and dl = 3 dφ a_φ. (Key point: Boundary of the patch)

2. What is the closed line integral of G?
   A. 30π
   B. 15π
   C. 10π
   D. 15π/2
   Answer: B. ∫ from 0 to π/2 of 10(3) dφ = 30(π/2) = 15π ≈ 47.12. (Key point: Line integral value)

3. With dS along a_r, which component of the curl matters, and what is it?
   A. (curl G)_r = 20 cosθ / r
   B. (curl G)_θ = -10 sinθ / r
   C. (curl G)_φ = 0, so the flux is zero
   D. (curl G)_r = 10 sinθ / r
   Answer: A. Only the a_r component of the curl contributes: (1/(r sinθ)) ∂(sinθ G_φ)/∂θ = 20 cosθ / r. (Key point: Curl component used)

4. What does evaluating the surface integral give, relative to the line integral?
   A. Twice the line integral
   B. Zero
   C. The same value 15π, which verifies the theorem
   D. Half the line integral
   Answer: C. 60 ∫ sinθ cosθ dθ ∫ dφ = 60(1/2)(π/2) = 15π, equal to the line integral. (Key point: Verification)

## Related in Electromagnetism

- [Verify Stokes' theorem, conical surface](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Averify-stokes-conical-surface) — Electromagnetism, chapter 1
- [Verify divergence theorem, D field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Averify-divergence-theorem-dr-theta) — Electromagnetism, chapter 1
- [Stokes' theorem and line integral of H](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Astokes-theorem-closed-line-integral) — Electromagnetism, chapter 1
- [State and prove Stokes' theorem](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Astokes-theorem-state-prove) — Electromagnetism, chapter 1, asked 3×
- [Verify divergence theorem, quarter hemisphere](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Averify-divergence-theorem-hemisphere) — Electromagnetism, chapter 1
- [Total current through surface, H field](https://www.bctengineer.com/question.md?key=semester_3%2Felectromagnetism%3Atotal-current-through-surface-h-rho) — Electromagnetism, chapter 3

## Related in other subjects

- [Evaluate using Stokes' theorem F=yi+x(1−2z)j−xyk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-stokes-theorem-f-yi) — Mathematics III, chapter 2, asked 2×
- [Evaluate using Stokes' theorem F=(2x−y)i−yz²j−y²zk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-stokes-theorem-f-2x) — Mathematics III, chapter 2
- [Evaluate using Stokes' theorem](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-stokes-theorem) — Mathematics III, chapter 2, asked 2×
- [Evaluate using Stokes' theorem](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-stokes-theorem-three) — Mathematics III, chapter 2, asked 2×