---
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-ten"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-ten
site: BCT Engineering
---

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

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

- Asked in: 2067 Magh
- 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-ten
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds where F = x i + y j + z k and S is the outside of the lateral surface of the circular cylinder x² + y² = a² between the planes z = 0 and z = 4.

## Short answer

Outward normal n = (x i + y j)/a gives F·n = a. Lateral area = 2πa × 4 = 8πa. Flux = a × 8πa = 8πa².

## Answer

S is the curved (lateral) surface of the cylinder x² + y² = a², 0 ≤ z ≤ 4, with outward normal n = (x i + y j)/a.

```text
F = x i + y j + z k
F·n = (x² + y²)/a = a²/a = a
```

On S the integrand is constant. The area of the lateral surface is circumference × height = 2πa × 4 = 8πa.

```text
∬_S F·n dS = a × 8πa = 8πa²
```

**Check with parameters:** x = a cos θ, y = a sin θ, dS = a dθ dz:

```text
∫₀⁴ ∫₀^2π a · a dθ dz = a² · 2π · 4 = 8πa²
```

**Answer: ∬_S F·n dS = 8πa².**

## Check yourself

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

1. The outward normal to the cylinder x^2+y^2=a^2 is:
   A. (x i + y j + z k)/a
   B. k
   C. (x i + y j)/a
   D. (x i - y j)/a
   Answer: C. It is radial in the xy-plane. (Key point: Normal)

2. F.n on the surface equals:
   A. a^2
   B. 1
   C. 0
   D. a
   Answer: D. (x^2 + y^2)/a = a^2/a = a. (Key point: F.n)

3. Lateral surface area is:
   A. 8 pi a
   B. 4 pi a^2
   C. 2 pi a
   D. 8 pi a^2
   Answer: A. Circumference 2 pi a times height 4. (Key point: Area)

4. The flux is:
   A. 8 pi a
   B. 8 pi a^2
   C. 4 pi a^2
   D. 2 pi a^2
   Answer: B. a * 8 pi a = 8 pi a^2. (Key point: Result)

## Related in Mathematics III

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## Related in other subjects

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