---
title: "Surface integral over a surface F=x²i+y²j+z²k"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:surface-integral-over-a-surface-f-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/surface-integral-over-a-surface-f-two
site: BCT Engineering
---

# Surface integral over a surface F=x²i+y²j+z²k

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

- Asked in: 2072 Chaitra, 5 marks; 2070 Chaitra, 5 marks; 2069 Asar, 5 marks (asked 3 times)
- Video: [Vector Calculus - Gauss Divergence Theorem | Example and Solution](https://www.youtube.com/watch?v=_PNsPatRNVA) — Dr.Gajendra Purohit
- 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds, where F = x² i + y² j + z² k and S is the surface of the plane x + y + z = 1 between the co-ordinate planes.

## Short answer

Normal vector is n = (i+j+k)/√3 and dS = √3 dxdy. With z = 1−x−y, the flux becomes ∬ (x²+y²+(1−x−y)²) dxdy over the triangle, which evaluates to 1/4.

## Answer

F = x² i + y² j + z² k, and S is the part of the plane x + y + z = 1 lying in the first octant (between the coordinate planes).

#### Normal and surface element
For φ = x + y + z, ∇φ = i + j + k, so n = (i + j + k)/√3. Projecting S onto the xy-plane (region R: x ≥ 0, y ≥ 0, x + y ≤ 1):

```text
dS = dx dy / |n·k| = √3 dx dy
```

#### F·n
```text
F·n = (x² + y² + z²)/√3 , with z = 1 − x − y
```

#### Integral
```text
∬_S F·n dS = ∬_R (x² + y² + (1−x−y)²) dx dy
```

Over the triangle R, each of the three terms gives the same value 1/12:
- ∬ x² dy dx = ∫₀¹ x²(1−x) dx = 1/3 − 1/4 = 1/12
- ∬ y² dx dy = 1/12 by symmetry
- ∬ (1−x−y)² dx dy = 1/12 by symmetry (it is the same integral in the third variable)

```text
∬_S F·n dS = 1/12 + 1/12 + 1/12 = 1/4
```

**Answer: ∬_S F·n dS = 1/4.**

#### Check by the divergence theorem
∇·F = 2(x + y + z). Over the tetrahedron x + y + z ≤ 1: ∭ (x + y + z) dV = 3 · (1/24) = 1/8, so ∭ ∇·F dV = 1/4. On the three coordinate faces F·n = 0 (for example on x = 0, F·(−i) = −x² = 0), so the flux through the slanted face is 1/4 ✓.

## Check yourself

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

1. Unit normal of x+y+z=1:
   A. i+j+k
   B. (i+j+k)/√3
   C. (i+j+k)/3
   D. k
   Answer: B. ∇φ/|∇φ|. (Key point: Normal)

2. Projecting onto xy-plane, dS =
   A. √3 dx dy
   B. dx dy/√3
   C. 3 dx dy
   D. dx dy
   Answer: A. 1/|n·k| = √3. (Key point: Surface element)

3. The integral becomes:
   A. ∬(x+y+z) dx dy
   B. ∬ dx dy
   C. ∬ x²y² dx dy
   D. ∬(x²+y²+(1−x−y)²) dx dy
   Answer: D. F·n dS = (x²+y²+z²) dx dy. (Key point: Integrand)

4. Value:
   A. 1/12
   B. 1/3
   C. 1/4
   D. 1
   Answer: C. Three terms of 1/12. (Key point: Result)

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