---
title: "Surface integral over a surface F=6zi−4j+yk"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:surface-integral-over-a-surface-f-five"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-five
site: BCT Engineering
---

# Surface integral over a surface F=6zi−4j+yk

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

- Asked in: 2075 Asoj, 5 marks
- Video: [F= (18zi - 12j+3yk) in first octants | Multivariable Calculus | engineering](https://www.youtube.com/watch?v=D4LXKyDqn9g) — Asif Concepts
- 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-five
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds where F = 6z i − 4 j + y k and S is the region of the plane 2x + 3y + 6z = 12 bounded in the first octant.

## Short answer

Project the plane onto the xy-plane: dS = (7/6) dxdy, F·n = (12−4x)/7. Then ∬ F·n dS = (1/6)∬ (12−4x) dxdy over the triangle 2x+3y≤12, x,y≥0, giving 8.

## Answer

S is the part of the plane 2x + 3y + 6z = 12 in the first octant (a triangle with vertices (6,0,0), (0,4,0), (0,0,2)).

```text
φ = 2x + 3y + 6z
∇φ = 2i + 3j + 6k,  |∇φ| = 7
n = (2i + 3j + 6k)/7   (outward, away from the origin)
F·n = (6z·2 − 4·3 + y·6)/7 = (12z − 12 + 6y)/7
```

Project S onto the xy-plane (region R: triangle 2x + 3y ≤ 12 with x, y ≥ 0): dS = dx dy/|n·k| = (7/6) dx dy. On S, z = (12 − 2x − 3y)/6, so 12z = 24 − 4x − 6y and

```text
F·n = (24 − 4x − 6y − 12 + 6y)/7 = (12 − 4x)/7
∬_S F·n dS = ∬_R (12 − 4x)/7 · (7/6) dx dy
           = (1/6) ∬_R (12 − 4x) dx dy
```

R: x from 0 to 6, y from 0 to (12 − 2x)/3.

```text
∫₀⁶ (12 − 4x)(12 − 2x)/3 dx = (1/3) ∫₀⁶ (144 − 72x + 8x²) dx
                            = (1/3)(864 − 1296 + 576)
                            = (1/3)(144) = 48
∬_S F·n dS = 48/6 = 8
```

**Answer: ∬_S F·n dS = 8.**

## Check yourself

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

1. n for 2x+3y+6z=12?
   A. (2i+3j+6k)/12
   B. (i+j+k)/3
   C. (2i+3j+6k)/7
   D. (2i+3j+6k)
   Answer: C. Gradient over magnitude 7. (Key point: Normal)

2. dS in terms of dx dy?
   A. (6/7) dx dy
   B. 7 dx dy
   C. dx dy
   D. (7/6) dx dy
   Answer: D. 1/|n.k|. (Key point: Area element)

3. F.n on S?
   A. (12-4x)/7
   B. (12+4x)/7
   C. (12z)/7
   D. 12-4x
   Answer: A. y cancels. (Key point: F.n)

4. Flux?
   A. 48
   B. 8
   C. 6
   D. 12
   Answer: B. 48/6. (Key point: Result)

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