---
title: "Flux through a surface F=(x+y²)i−2xj+2yzk"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:flux-through-a-surface-f-x"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/flux-through-a-surface-f-x
site: BCT Engineering
---

# Flux through a surface F=(x+y²)i−2xj+2yzk

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

- Asked in: 2079 Baishakh, 5 marks
- 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/flux-through-a-surface-f-x
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate the flux of F = (x + y²) i − 2x j + 2yz k over the surface of the plane 2x + y + 2z = 6 lying in the first octant.

## Short answer

Using projection onto the xy-plane, F·n dS = (6y−2xy) dx dy. Integrating over 0≤x≤3, 0≤y≤6−2x yields flux = 81.

## Answer

Flux = ∬_S F·n dS over the part of the plane 2x + y + 2z = 6 in the first octant (triangle with vertices (3,0,0), (0,6,0), (0,0,3)).

```text
∇φ = 2i + j + 2k,  |∇φ| = 3,  n = (2i + j + 2k)/3   (away from origin)
F = (x + y²) i − 2x j + 2yz k
F·n = [2(x + y²) − 2x + 4yz]/3 = (2y² + 4yz)/3
```

Project on the xy-plane: |n·k| = 2/3, so dS = (3/2) dx dy. Then

```text
F·n dS = (y² + 2yz) dx dy
```

On the plane z = (6 − 2x − y)/2, so 2yz = y(6 − 2x − y) and y² + 2yz = 6y − 2xy.

Limits: x from 0 to 3, y from 0 to 6 − 2x.

```text
∫₀^(6−2x) (6y − 2xy) dy = (3 − x)(6 − 2x)² = 4(3 − x)³
∫₀³ 4(3 − x)³ dx = 4 · 81/4 = 81
```

**Answer: Flux = 81.**

## Check yourself

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

1. n for 2x+y+2z=6:
   A. (2i+j+2k)
   B. (i+j+k)/3
   C. (2i+j+2k)/3
   D. (2i+j+2k)/6
   Answer: C. |grad|=3. (Key point: Normal)

2. dS =
   A. (2/3) dx dy
   B. 3 dx dy
   C. dx dy
   D. (3/2) dx dy
   Answer: D. n.k = 2/3. (Key point: dS)

3. F.n dS reduces to:
   A. (y^2+2yz) dx dy
   B. (x+y^2)dx dy
   C. (2y^2+4yz) dx dy
   D. 2yz dx dy
   Answer: A. (2y^2+4yz)/3 * 3/2. (Key point: Integrand)

4. Flux:
   A. 27
   B. 81
   C. 54
   D. 162
   Answer: B. 4 * 81/4. (Key point: Result)

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