---
title: "Evaluate using Gauss' divergence theorem F=x²yi+xy²j+2xyzk"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:evaluate-using-gauss-divergence-theorem-f"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/evaluate-using-gauss-divergence-theorem-f
site: BCT Engineering
---

# Evaluate using Gauss' divergence theorem F=x²yi+xy²j+2xyzk

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

- Asked in: 2083 Baishakh, 5 marks; 2082 Baishakh, 5 marks; 2075 Asoj, 5 marks; 2071 Shrawan, 5 marks (asked 4 times)
- Video: [Applications of Gauss Divergence Theorem on the tetrahedron / problem](https://www.youtube.com/watch?v=DYH4JPzoXRg) — 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/evaluate-using-gauss-divergence-theorem-f
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Using Gauss' divergence theorem, evaluate ∬_S F·n ds, where F = x²y i + xy² j + 2xyz k and S is the surface bounded by the planes x = 0, y = 0, z = 0, x + 2y + z = 2.

## Short answer

Divergence ∇·F = 6xy. Integrating over the tetrahedron bounded by the coordinate planes and x+2y+z=2 with limits z = 0→2−x−2y, y = 0→(2−x)/2, x = 0→2 yields a flux of 2/5.

## Answer

**Gauss' divergence theorem:** ∬_S F·n dS = ∭_V (∇·F) dV for a closed surface S enclosing V.

F = x²y i + xy² j + 2xyz k

#### Divergence
```text
∇·F = ∂(x²y)/∂x + ∂(xy²)/∂y + ∂(2xyz)/∂z = 2xy + 2xy + 2xy = 6xy
```

#### Region V
The tetrahedron bounded by x = 0, y = 0, z = 0 and x + 2y + z = 2. Its intercepts are x = 2, y = 1, z = 2. Limits:

```text
z : 0 → 2 − x − 2y
y : 0 → (2 − x)/2
x : 0 → 2
```

#### Evaluate
```text
∭ 6xy dV = ∫₀² ∫₀^{(2−x)/2} 6xy (2 − x − 2y) dy dx
```

Let k = 2 − x. Inner integral:

```text
∫₀^{k/2} 6xy(k − 2y) dy = 6x [ k y²/2 − 2y³/3 ]₀^{k/2}
                        = 6x ( k³/8 − k³/12 ) = x k³/4
```

Outer integral (put u = 2 − x, x = 2 − u):

```text
∫₀² x(2−x)³/4 dx = (1/4) ∫₀² (2 − u) u³ du
                 = (1/4) [ u⁴/2 − u⁵/5 ]₀² = (1/4)(8 − 32/5) = (1/4)(8/5) = 2/5
```

**Answer: ∬_S F·n dS = 2/5.**

## Check yourself

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

1. Gauss' theorem converts ∬F·n dS to:
   A. ∭∇×F dV
   B. ∭∇·F dV
   C. ∮F·dr
   D. ∬∇·F dS
   Answer: B. Surface integral over closed surface equals volume integral of divergence. (Key point: Theorem)

2. ∇·F for F=x²y i+xy² j+2xyz k:
   A. 6xy
   B. 2xy
   C. x²+y²
   D. 0
   Answer: A. 2xy+2xy+2xy. (Key point: Divergence)

3. The tetrahedron has intercepts:
   A. x=2, y=2, z=1
   B. x=1, y=1, z=1
   C. x=1, y=2, z=2
   D. x=2, y=1, z=2
   Answer: D. From x+2y+z=2. (Key point: Region)

4. The value is:
   A. 1/5
   B. 3/5
   C. 2/5
   D. 2
   Answer: C. Evaluating the triple integral gives 2/5. (Key point: Result)

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