---
title: "Flux through a surface F=3xi+3yj+zk"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:flux-through-a-surface-f-3xi"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/flux-through-a-surface-f-3xi
site: BCT Engineering
---

# Flux through a surface F=3xi+3yj+zk

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

- Asked in: 2068 Baishakh, 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/flux-through-a-surface-f-3xi
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

If S is the part of the surface z = 9 − x² − y² with z ≥ 0 and F = 3x i + 3y j + z k, find the flux of F through S.

## Short answer

Close the paraboloid with the base disc and apply Gauss theorem. div F = 7, volume under z=9−x²−y² is 81π/2, total flux 567π/2. Base flux is zero, so flux through S = 567π/2.

## Answer

S is the open paraboloid cap z = 9 − x² − y² (z ≥ 0). Its boundary is the circle x² + y² = 9 in the plane z = 0. Close the surface with the disc S₁: x² + y² ≤ 9, z = 0, and use Gauss' theorem.

```text
F = 3x i + 3y j + z k
div F = 3 + 3 + 1 = 7
```

#### Volume under the paraboloid
```text
V = ∬ (9 − r²) r dr dθ = 2π [9r²/2 − r⁴/4]₀³ = 2π (81/2 − 81/4) = 81π/2
```

#### Total flux through the closed surface
```text
∬_(S+S₁) F·n dS = 7V = 7 × 81π/2 = 567π/2
```

#### Flux through the base
On S₁: z = 0, n = −k, so F·n = −z = 0. Flux through S₁ = 0.

```text
Flux through S = 567π/2 − 0
```

**Answer: Flux through S = 567π/2 ≈ 890.6.**

## Check yourself

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

1. Why is the disc S1 (x^2+y^2<=9, z=0) added to the paraboloid cap?
   A. To make a closed surface so Gauss' theorem can be applied
   B. To make the surface flat so projection is easier
   C. To remove the boundary circle from the answer
   D. Because the cap alone has zero divergence
   Answer: A. Gauss' theorem needs a closed surface, so the base disc closes the cap and its flux is subtracted afterwards. (Key point: Closing the surface)

2. What is div F for F = 3x i + 3y j + z k?
   A. 3
   B. 7
   C. 6
   D. 9
   Answer: B. div F = 3 + 3 + 1 = 7. (Key point: Divergence)

3. What is the volume under z = 9 - r^2 above z = 0?
   A. 81pi
   B. 9pi
   C. 81pi/2
   D. 81pi/4
   Answer: C. V = 2pi[9r^2/2 - r^4/4] from 0 to 3 = 81pi/2. (Key point: Volume)

4. Why is the flux through the base disc S1 zero?
   A. Because the disc has zero area
   B. Because div F is zero on the disc
   C. Because n is perpendicular to the disc
   D. On z = 0 with n = -k, F.n = -z = 0
   Answer: D. F.n = -z, which vanishes at z = 0, so the flux through S1 is 0 and flux through S = 567pi/2. (Key point: Base flux)

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