---
title: "Verify divergence theorem, quarter hemisphere"
subject: "Electromagnetism"
chapter: 1
key: "semester_3/electromagnetism:verify-divergence-theorem-hemisphere"
source: https://www.bctengineer.com/q/semester_3/electromagnetism/verify-divergence-theorem-hemisphere
site: BCT Engineering
---

# Verify divergence theorem, quarter hemisphere

Electromagnetism · Chapter 1: Introduction

- Asked in: 2069 Chaitra, 6 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/electromagnetism/verify-divergence-theorem-hemisphere
- Whole chapter: https://www.bctengineer.com/md/semester_3/electromagnetism/chapter-1

## Question

Verify the divergence theorem (evaluate both sides of the divergence theorem) for the function A = r² a_r + r sinθ cosφ a_θ over the surface of a quarter of a hemisphere defined by 0 < r < 3, 0 < φ < π/2, 0 < θ < π/2.

## Short answer

The volume integral of ∇·A over the region is 81π/2+9, matching the total flux through the bounding surfaces, verifying the theorem.

## Answer

**Divergence theorem:** ∮ A · dS = ∫ (∇ · A) dv.

**Given:** A = r² a_r + r sinθ cosφ a_θ. Region: 0 < r < 3, 0 < θ < π/2, 0 < φ < π/2 (one-eighth of a sphere).

#### Volume integral

```text
∇ · A = (1/r²) ∂(r² A_r)/∂r + (1/(r sinθ)) ∂(sinθ A_θ)/∂θ + (1/(r sinθ)) ∂A_φ/∂φ

(1/r²) ∂(r⁴)/∂r = 4r
(1/(r sinθ)) ∂(r sin²θ cosφ)/∂θ = (1/(r sinθ)) (2r sinθ cosθ cosφ) = 2 cosθ cosφ

∇ · A = 4r + 2 cosθ cosφ
```

```text
∫ (∇ · A) dv = ∫₀^(π/2) ∫₀^(π/2) ∫₀³ (4r + 2cosθ cosφ) r² sinθ dr dθ dφ

Term 1: 4 ∫₀³ r³ dr · ∫₀^(π/2) sinθ dθ · ∫₀^(π/2) dφ = 4 (81/4)(1)(π/2) = 81π/2 = 127.23
Term 2: 2 ∫₀³ r² dr · ∫₀^(π/2) sinθ cosθ dθ · ∫₀^(π/2) cosφ dφ = 2 (9)(1/2)(1) = 9

Total = 81π/2 + 9 = 136.23
```

#### Surface integral over the four faces

```text
1. Curved face r = 3 (dS = r² sinθ dθ dφ a_r):
   Ψ₁ = ∫∫ (3²)(3²) sinθ dθ dφ = 81 (1)(π/2) = 81π/2 = 127.23

2. Face θ = π/2 (outward normal +a_θ, dS = r sinθ dr dφ a_θ):
   A_θ = r cosφ at θ = π/2
   Ψ₂ = ∫₀^(π/2)∫₀³ (r cosφ)(r) dr dφ = (9)(1) = 9

3. Faces φ = 0 and φ = π/2 (normal ±a_φ): A has no a_φ component → 0

4. The face r = 0 is a point, θ = 0 is the z-axis line → 0
```

```text
Ψ = 127.23 + 9 = 136.23
```

**Answer:** Both sides equal **81π/2 + 9 ≈ 136.2**. The divergence theorem is verified.

## Check yourself

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

1. What is ∇·A for A = r^2 a_r + r sinθ cosφ a_θ?
   A. 4r + 2 cosθ cosφ
   B. 2r + cosθ
   C. 4r
   D. 4r - 2cosθ cosφ
   Answer: A. The radial part gives (1/r²)∂(r⁴)/∂r = 4r and the θ part gives 2 cosθ cosφ. (Key point: Divergence)

2. What is the volume integral of ∇·A?
   A. 81π/2
   B. 81π/2 + 9 ≈ 136.2
   C. 9
   D. 81π
   Answer: B. The 4r term gives 81π/2 = 127.23 and the 2cosθ cosφ term gives 9, totalling 136.23. (Key point: Volume integral value)

3. Which faces carry flux in the surface integral?
   A. Only the φ faces
   B. The curved face r = 3 (127.23) and the face θ = π/2 (9)
   C. r = 0 and θ = 0
   D. All faces equally
   Answer: B. The curved face gives 81π/2 and θ = π/2 gives 9. The φ faces get zero because A has no a_φ component, and r = 0 and θ = 0 have no area. (Key point: Faces contributing flux)

4. Why is the outward normal on face θ = π/2 equal to +a_θ?
   A. θ increases outward there
   B. Because A_θ is positive
   C. Because r is outward
   D. Because φ is constant
   Answer: A. The region is 0 < θ < π/2, so at θ = π/2 the outward direction is the direction of increasing θ, +a_θ. (Key point: Face θ = π/2)

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