---
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-seven"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-seven
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: 2074 Asoj, 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-seven
- 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 and S is the region of the plane 2x + 2y + z = 6 bounded in the first quadrant.

## Short answer

Project onto xy-plane: normal n = (2i+2j+k)/3, dS = 3 dx dy. Substitute z = 6−2x−2y into F·n to get (2xy−2x²−x−2y+6) dx dy. Integrate over x≥0, y≥0, x+y≤3 yields 27/4.

## Answer

"First quadrant" here means the first octant: S is the part of the plane 2x + 2y + z = 6 with x, y, z ≥ 0 (triangle with vertices (3,0,0), (0,3,0), (0,0,6)).

```text
φ = 2x + 2y + z,   ∇φ = 2i + 2j + k,  |∇φ| = 3
n = (2i + 2j + k)/3   (upward / away from origin)
F = xy i − x² j + (x + z) k
F·n = (2xy − 2x² + x + z)/3
```

Project onto the xy-plane: dS = dx dy/|n·k| = 3 dx dy. On S, z = 6 − 2x − 2y.

```text
F·n dS = (2xy − 2x² + x + 6 − 2x − 2y) dx dy
       = (2xy − 2x² − x − 2y + 6) dx dy
```

Region R: triangle x ≥ 0, y ≥ 0, x + y ≤ 3. Integrate y from 0 to 3 − x, then x from 0 to 3.

Let s = 3 − x. Integrating (2xy − 2x² − x − 2y + 6) over y from 0 to s:

```text
Inner = x s² + (6 − x − 2x²) s − s²
      = s [x s + 6 − x − 2x² − s]
      = s [3x − x² + 6 − x − 2x² − 3 + x]
      = 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) = 3 (9/4) = 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. 'First quadrant' means?
   A. x,y>=0 only
   B. xy-plane
   C. First octant, x,y,z >= 0
   D. z=0
   Answer: C. Triangle. (Key point: Interpretation)

2. n?
   A. (i+j+k)/3
   B. (2i+2j+k)
   C. (2i+2j+k)/9
   D. (2i+2j+k)/3
   Answer: D. Magnitude 3. (Key point: Normal)

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

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

## Close variants set in other exams

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