---
title: "Surface integral over a surface F=xyi−x²j+(x+z)k"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:surface-integral-over-a-surface-f-twelve"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-twelve
site: BCT Engineering
---

# Surface integral over a surface F=xyi−x²j+(x+z)k

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

- Asked in: 2070 Asar, 5 marks
- 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-twelve
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds where F = xy i − x² j + (x + z) k, S is the portion of the plane 2x + 2y + z = 6 included in the first octant.

## Short answer

Project onto the xy-plane: F·n dS = (2xy−2x²−x−2y+6) dx dy. Integrating over the triangular region x,y≥0, x+y≤3 gives 27/4 (6.75).

## Answer

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

```text
∇φ = 2i + 2j + k,  |∇φ| = 3,  n = (2i + 2j + k)/3
F = xy i − x² j + (x + z) k
F·n = (2xy − 2x² + x + z)/3
dS = dx dy/|n·k| = 3 dx dy
```

On S, z = 6 − 2x − 2y, so F·n dS = (2xy − 2x² + x + 6 − 2x − 2y) dx dy = (2xy − 2x² − x − 2y + 6) dx dy.

The projection R is the triangle x ≥ 0, y ≥ 0, x + y ≤ 3. With s = 3 − x:

```text
∫₀^s (2xy − 2x² − x − 2y + 6) dy = x s² + (6 − x − 2x²) s − s²
                                 = 3 s (1 + x − x²)

∫₀³ 3(3 − x)(1 + x − x²) dx = 3 ∫₀³ (3 + 2x − 4x² + x³) dx
                            = 3 [3x + x² − 4x³/3 + x⁴/4]₀³
                            = 3 (9 + 9 − 36 + 81/4) = 27/4
```

**Answer: ∬_S F·n dS = 27/4 = 6.75.**

## Check yourself

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

1. n for 2x+2y+z=6:
   A. (2i+2j+k)
   B. (i+j+k)/3
   C. (2i+2j+k)/3
   D. (2i+2j+k)/9
   Answer: C. |grad|=3. (Key point: Normal)

2. dS =
   A. dx dy/3
   B. dx dy
   C. 9 dx dy
   D. 3 dx dy
   Answer: D. n.k = 1/3. (Key point: dS)

3. Projection is triangle:
   A. x,y>=0, x+y<=3
   B. x+y<=6
   C. x,y<=3
   D. x+y<=2
   Answer: A. Set z=0. (Key point: Region)

4. Value:
   A. 27
   B. 27/4
   C. 9/4
   D. 6
   Answer: B. Computed integral. (Key point: Result)

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