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

# Surface integral over a surface F=yzi+zxj+xyk

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

- Asked in: 2078 Bhadra, 5 marks; 2075 Chaitra, 5 marks; 2073 Shrawan, 5 marks; 2072 Kartik, 5 marks; 2070 Chaitra, 5 marks (asked 5 times)
- Video: [If F= yzi+zxj+xyk evaluate integral F.Nds over surface x^2+y^2+z^2=1 in first octact](https://www.youtube.com/watch?v=ZaNM_IvULiA) — Sudhakar Maths academy
- 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
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∬_S F·n ds where F = yz i + zx j + xy k where S is the surface of the sphere x² + y² + z² = 1 in the first octant.

## Short answer

With unit normal n = (x i + y j + z k), F·n = 3xyz. Using spherical coordinates x = sin φ cos θ, y = sin φ sin θ, z = cos φ and dS = sin φ dφ dθ, the integral over the first octant gives 3/8.

## Answer

Here F = yz i + zx j + xy k and S is the part of the unit sphere x² + y² + z² = 1 with x, y, z ≥ 0 (curved surface only).

#### Unit normal
For the sphere φ = x² + y² + z² = 1, n = ∇φ/|∇φ| = (x i + y j + z k) (since |∇φ| = 2 and the radius is 1).

#### F·n
```text
F·n = (yz)(x) + (zx)(y) + (xy)(z) = 3xyz
```

#### Parametrize the surface
x = sin φ cos θ, y = sin φ sin θ, z = cos φ, with 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ π/2, and dS = sin φ dφ dθ.

```text
xyz = sin²φ cos φ · sin θ cos θ
∬ F·n dS = 3 ∫₀^{π/2} ∫₀^{π/2} sin²φ cos φ sin θ cos θ · sin φ dφ dθ
         = 3 · [∫₀^{π/2} sin³φ cos φ dφ] · [∫₀^{π/2} sin θ cos θ dθ]
         = 3 · (1/4) · (1/2)
         = 3/8
```

**Answer: ∬ F·n dS = 3/8.**

#### Check by the divergence theorem
div F = ∂(yz)/∂x + ∂(zx)/∂y + ∂(xy)/∂z = 0, so the flux through the closed octant surface is 0. The flux out through the three coordinate quarter-discs is, for each, −∬ yz dy dz (or its analogue) = −1/8. Hence the curved part carries +3/8 ✓.

## Check yourself

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

1. For sphere x²+y²+z²=1, n is:
   A. i+j+k
   B. xi+yj+zk
   C. (xi+yj+zk)/3
   D. -(xi+yj+zk)
   Answer: B. ∇φ/|∇φ| = (2xi+2yj+2zk)/2. (Key point: Unit normal)

2. F·n for F=yzi+zxj+xyk:
   A. 3xyz
   B. xyz
   C. x+y+z
   D. 0
   Answer: A. Each term gives xyz. (Key point: F·n)

3. dS on the unit sphere is:
   A. dφ dθ
   B. cos φ dφ dθ
   C. sin²φ dφ dθ
   D. sin φ dφ dθ
   Answer: D. For r = 1. (Key point: Parametrisation)

4. The flux is:
   A. 3/4
   B. 1/8
   C. 3/8
   D. 0
   Answer: C. 3·(1/4)·(1/2) = 3/8. (Key point: Result)

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