---
title: "Verify Gauss' divergence theorem F=yi+xj+z²k"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:verify-gauss-divergence-theorem-f-yi"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-gauss-divergence-theorem-f-yi
site: BCT Engineering
---

# Verify Gauss' divergence theorem F=yi+xj+z²k

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

- Asked in: 2067 Asar
- Video: [Gauss divergence - F=4xi−2y^2j+z^2k and the region bounded by x^2+y^2=4, z= 0 and z=3](https://www.youtube.com/watch?v=p9jM8ZmzCdM) — Bright Concept
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-gauss-divergence-theorem-f-yi
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Verify Gauss divergence theorem for F = y i + x j + z² k over the region bounded by x² + y² = a², z = 0 and z = h.

## Short answer

div F = 2z, volume integral = πa²h². Surface flux: top πa²h², bottom 0, curved surface 0. Sum equals volume integral, verifying the theorem.

## Answer

**Gauss' theorem:** ∬_S F·n dS = ∭_V div F dV, with F = y i + x j + z² k and V the cylinder x² + y² ≤ a², 0 ≤ z ≤ h.

#### Volume integral
```text
div F = 0 + 0 + 2z = 2z
∭ 2z dV = ∫₀^h 2z (πa²) dz = πa² h²
```
(the cross-section at height z has area πa²).

#### Surface integral
- **Top** z = h, n = k: F·n = z² = h². Flux = h² · πa² = πa²h².
- **Bottom** z = 0, n = −k: F·n = −z² = 0. Flux = 0.
- **Curved surface**: n = (x i + y j)/a, so F·n = (xy + yx)/a = 2xy/a. With x = a cos θ, y = a sin θ, dS = a dθ dz:

```text
∫₀^h ∫₀^2π (2a² sin θ cos θ / a) a dθ dz = a² h ∫₀^2π sin 2θ dθ = 0
```

Total surface flux = πa²h² + 0 + 0 = πa²h², equal to the volume integral.

So **Gauss' divergence theorem is verified.**

## Check yourself

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

1. Gauss' divergence theorem states:
   A. Line integral of F equals surface integral of curl F
   B. Volume integral of F equals surface integral of div F
   C. Surface integral of F.n equals volume integral of div F
   D. Surface integral of F.n equals line integral of div F
   Answer: C. Closed surface flux = triple integral of div F. (Key point: Theorem)

2. What is the volume integral?
   A. 2 pi a^2 h
   B. pi a^2 h
   C. pi a h^2
   D. pi a^2 h^2
   Answer: D. div F = 2z; integral = integral of 2z * pi a^2 dz = pi a^2 h^2. (Key point: Volume side)

3. Why is the flux through the curved surface zero?
   A. F.n = 2xy/a gives a sin 2theta integral over 0 to 2pi, which is 0
   B. n is perpendicular to F
   C. dS is zero on the sides
   D. z = 0 on the curved side
   Answer: A. The integral of sin 2theta over a full period is 0. (Key point: Curved surface)

4. Fluxes through top and bottom are:
   A. 0 and pi a^2 h^2
   B. pi a^2 h^2 and 0
   C. pi a^2 h^2 each
   D. h^2 and 0
   Answer: B. Top F.n = h^2 times area pi a^2; bottom F.n = -z^2 = 0. (Key point: Top and bottom)

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