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

# Surface integral over a surface F=xi−yj+zk

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

- Asked in: 2074 Chaitra, 5 marks
- Video: [Evaluation of surface integral over the cylinder in first octant](https://www.youtube.com/watch?v=0UguqZ2SEBo) — TEAM Education
- 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-six
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·ds where F = x i − y j + z k and S is the surface of the cylinder x² + y² = a², 0 < z < b.

## Short answer

For the curved surface, outward unit normal is (x i + y j)/a, so F·n = (x²−y²)/a. Using x = a cosθ, y = a sinθ, dS = a dθ dz, the integral becomes a²∫₀^b∫₀^{2π} cos2θ dθ dz = 0. If the closed cylinder is intended, the total flux is πa²b.

## Answer

S is the curved surface of the cylinder x² + y² = a², 0 < z < b. For this surface the outward unit normal is n = (x i + y j)/a (no k component).

```text
F = x i − y j + z k
F·n = (x² − y²)/a
```

Use x = a cos θ, y = a sin θ, dS = a dθ dz, with 0 ≤ θ ≤ 2π, 0 ≤ z ≤ b.

```text
∬_S F·n dS = ∫₀^b ∫₀^2π [(a² cos²θ − a² sin²θ)/a] · a dθ dz
           = a² ∫₀^b ∫₀^2π cos 2θ dθ dz
           = a² · b · [sin 2θ / 2]₀^2π
           = 0
```

**Answer: ∬_S F·n dS = 0** for the curved surface.

**Remark.** If the two circular ends are also counted (closed cylinder), div F = 1 − 1 + 1 = 1, so by Gauss' theorem the total flux is the volume = πa²b. This comes entirely from the top (z = b): F·k = b over area πa²; the bottom gives 0 since z = 0.

## Check yourself

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

1. n for cylinder?
   A. (x i + y j + z k)/a
   B. k
   C. (x i + y j)/a
   D. (i+j)/a
   Answer: C. No k component. (Key point: Normal)

2. F.n?
   A. (x^2+y^2)/a
   B. z/a
   C. 0
   D. (x^2 - y^2)/a
   Answer: D. Dot product. (Key point: F.n)

3. Flux of curved surface?
   A. 0
   B. pi a^2 b
   C. a^2 b
   D. 2pi a
   Answer: A. cos 2θ integrates to zero. (Key point: Result)

4. With end discs, total flux?
   A. 0
   B. pi a^2 b
   C. pi a b
   D. 2 pi a^2 b
   Answer: B. div F = 1, volume. (Key point: Closed cylinder)

## Related in Mathematics III

- [Surface integral over a surface F=xi+yj+zk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asurface-integral-over-a-surface-f-ten) — Mathematics III, chapter 2
- [Surface integral over a surface F=xi+yj+zk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asurface-integral-over-a-surface-f-eleven) — Mathematics III, chapter 2
- [Evaluate using Gauss' divergence theorem F=xi+yj+zk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-gauss-divergence-theorem-f-eight) — Mathematics III, chapter 2, asked 4×
- [Evaluate using Gauss' divergence theorem F=xi+yj+z²k](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-gauss-divergence-theorem-f-six) — Mathematics III, chapter 2, asked 6×
- [Evaluate using Gauss' divergence theorem F=xi−yj+(z²−1)k](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Aevaluate-using-gauss-divergence-theorem-f-two) — Mathematics III, chapter 2, asked 3×
- [Surface integral over a surface F=3i+xj−yzk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Asurface-integral-over-a-surface-f-eight) — Mathematics III, chapter 2

## Related in other subjects

- [Triple integral (2x+y) over cylinder region](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Atriple-integral-2x-y-over) — Mathematics, chapter 2