---
title: "Surface integral over a surface F=(x−2z)i+(x+3y+z)j…"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:surface-integral-over-a-surface-f-nine"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-nine
site: BCT Engineering
---

# Surface integral over a surface F=(x−2z)i+(x+3y+z)j…

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

- Asked in: 2067 Asar
- Video: [Evaluation of Surface Integral in a plane 2x+y+2z=6 in the first octant](https://www.youtube.com/watch?v=xzL5WazCySo) — 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-nine
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds where S is the upper side of the triangle with vertices (1,0,0), (0,1,0), (0,0,1) and F = (x − 2z) i + (x + 3y + z) j + (5x + y) k.

## Short answer

On plane x+y+z=1, n = (i+j+k)/√3, F·n dS = (8x+5y−1) dx dy over the triangle x,y≥0, x+y≤1. Integration yields 5/3.

## Answer

S is the triangle on the plane x + y + z = 1 with vertices (1,0,0), (0,1,0), (0,0,1). The upper side has n = (i + j + k)/√3.

```text
F = (x − 2z) i + (x + 3y + z) j + (5x + y) k
F·n = [(x − 2z) + (x + 3y + z) + (5x + y)]/√3
    = (7x + 4y − z)/√3
```

Project onto the xy-plane: the region R is the triangle x ≥ 0, y ≥ 0, x + y ≤ 1, and dS = dx dy/|n·k| = √3 dx dy. On S, z = 1 − x − y.

```text
F·n dS = (7x + 4y − (1 − x − y)) dx dy = (8x + 5y − 1) dx dy

∬_R (8x + 5y − 1) dx dy = ∫₀¹ ∫₀^(1−x) (8x + 5y − 1) dy dx
```

Using area(R) = 1/2, ∬ x dA = 1/6, ∬ y dA = 1/6:

```text
= 8(1/6) + 5(1/6) − 1/2 = 13/6 − 3/6 = 10/6 = 5/3
```

**Answer: ∬_S F·n dS = 5/3.**

## Check yourself

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

1. What is the upper unit normal of the plane x+y+z=1?
   A. i + j + k
   B. (i + j - k)/sqrt(3)
   C. (i + j + k)/sqrt(3)
   D. (i + j + k)/3
   Answer: C. grad phi = i+j+k with magnitude sqrt(3). (Key point: Unit normal)

2. How does dS relate to dx dy here?
   A. dS = dx dy/sqrt(3)
   B. dS = dx dy
   C. dS = 3 dx dy
   D. dS = sqrt(3) dx dy
   Answer: D. dS = dx dy/|n.k| and n.k = 1/sqrt(3). (Key point: dS)

3. After substituting z = 1-x-y, F.n dS becomes:
   A. (8x + 5y - 1) dx dy
   B. (7x + 4y - 1) dx dy
   C. (6x + 3y + 1) dx dy
   D. (8x + 5y + 1) dx dy
   Answer: A. 7x + 4y - (1-x-y) = 8x + 5y - 1. (Key point: Integrand)

4. Using area 1/2 and integrals of x and y equal to 1/6, the result is:
   A. 13/6
   B. 5/3
   C. 1/2
   D. 10/3
   Answer: B. 8/6 + 5/6 - 1/2 = 10/6 = 5/3. (Key point: Result)

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