---
title: "Surface integral over a surface F=4xi−2y²j+z²k"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:surface-integral-over-a-surface-f-three"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-three
site: BCT Engineering
---

# Surface integral over a surface F=4xi−2y²j+z²k

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

- Asked in: 2083 Baishakh, 5 marks; 2076 Chaitra, 5 marks (asked 2 times)
- 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/surface-integral-over-a-surface-f-three
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds, where F = 4x i − 2y² j + z² k taken over the region bounded by the cylinder x² + y² = 4 and the planes z = 0, z = 3.

## Short answer

Using Gauss divergence theorem, ∇·F = 4 − 4y + 2z. Volume integral in cylindrical coordinates over cylinder radius 2, height 3 gives 84π.

## Answer

F = 4x i − 2y² j + z² k, taken over the closed surface bounded by the cylinder x² + y² = 4 and the planes z = 0, z = 3. Use Gauss' divergence theorem: ∬_S F·n dS = ∭_V ∇·F dV.

#### Divergence
```text
∇·F = ∂(4x)/∂x + ∂(−2y²)/∂y + ∂(z²)/∂z = 4 − 4y + 2z
```

#### Volume integral (cylindrical coordinates)
x = r cos θ, y = r sin θ, dV = r dr dθ dz; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 3.

```text
∭ (4 − 4y + 2z) dV = ∭ 4 dV − 4∭ y dV + 2∭ z dV
```

- ∭ 4 dV = 4 × (volume) = 4 × π(2)²(3) = 48π
- ∭ y dV = 0, because it contains ∫₀^{2π} sin θ dθ = 0
- 2∭ z dV = 2 · (∫₀³ z dz)(∫₀^{2π} dθ ∫₀² r dr) = 2 · (9/2)(4π) = 36π

```text
Total = 48π − 0 + 36π = 84π
```

**Answer: ∬_S F·n dS = 84π.**

## Check yourself

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

1. Why can the surface integral over the cylinder with planes z = 0, z = 3 be done with the divergence theorem?
   A. The surface is closed (curved side plus two end discs)
   B. F is irrotational
   C. The surface is a plane
   D. F has zero divergence
   Answer: A. Gauss' theorem needs a closed surface enclosing a volume. (Key point: Method)

2. What is div F for F = 4x i - 2y^2 j + z^2 k?
   A. 4 - 2y + 2z
   B. 4 - 4y + 2z
   C. 4 + 4y + 2z
   D. 4 - 4y + z
   Answer: B. d(4x)/dx = 4, d(-2y^2)/dy = -4y, d(z^2)/dz = 2z. (Key point: Divergence)

3. Why does the integral of y over the cylinder volume vanish?
   A. Because y = 0 on the cylinder
   B. Because the cylinder has zero volume
   C. It contains the integral of sin(theta) over 0 to 2pi, which is 0
   D. Because z-limits cancel it
   Answer: C. y = r sin(theta), and sin integrates to zero over a full turn. (Key point: Vanishing term)

4. What is the total?
   A. 48pi - 36pi = 12pi
   B. 48pi
   C. 36pi
   D. 48pi + 36pi = 84pi
   Answer: D. The constant term gives 4 x volume = 48pi, the z term gives 36pi, the y term gives 0. (Key point: Final value)

## Related in Mathematics III

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