---
title: "Evaluate using Stokes' theorem F=yi+x(1−2z)j−xyk"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:evaluate-using-stokes-theorem-f-yi"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/evaluate-using-stokes-theorem-f-yi
site: BCT Engineering
---

# Evaluate using Stokes' theorem F=yi+x(1−2z)j−xyk

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

- Asked in: 2081 Bhadra, 5 marks; 2081 Baishakh, 5 marks (asked 2 times)
- Video: [use stoke theorem int curlf. nds , f=yI+(x-2xz)J -xyK   and s is sphere x^2+y^2+z^2=1](https://www.youtube.com/watch?v=iFh5zPLH9AI) — maths solution advance
- 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-stokes-theorem-f-yi
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Apply Stokes' theorem to evaluate ∬_S (∇×F)·n ds, where F = y i + x(1 − 2z) j − xy k and S is the surface of the sphere x² + y² + z² = a² above the xy-plane.

## Short answer

By Stokes’ theorem, surface integral equals line integral around boundary circle z=0. On the circle, F·dr = d(xy), so ∮ F·dr = 0; thus the surface integral is zero.

## Answer

**Stokes' theorem:** ∬_S (∇×F)·n dS = ∮_C F·dr, where C is the boundary of S.

F = y i + x(1 − 2z) j − xy k. S is the upper hemisphere x² + y² + z² = a², z ≥ 0, whose boundary C is the circle x² + y² = a², z = 0.

#### Line integral on C
On C, z = 0 and dz = 0, so F·dr = y dx + x dy = d(xy). With x = a cos t, y = a sin t, 0 ≤ t ≤ 2π:

```text
∮_C (y dx + x dy) = [xy]  taken once round the closed curve = 0
```

Explicitly: ∫₀^{2π} (−a² sin²t + a² cos²t) dt = a² ∫₀^{2π} cos 2t dt = 0.

#### Check by direct surface integral
```text
∇×F = i(−x + 2x) + j(0 + y) + k((1 − 2z) − 1) = x i + y j − 2z k
```
On the sphere n = (x i + y j + z k)/a, so

```text
(∇×F)·n = (x² + y² − 2z²)/a
```
With x = a sin φ cos θ, y = a sin φ sin θ, z = a cos φ, we get (∇×F)·n = a(sin²φ − 2cos²φ) and dS = a² sin φ dφ dθ:

```text
∬ = a³ ∫₀^{2π} ∫₀^{π/2} (sin²φ − 2cos²φ) sin φ dφ dθ
  = a³ · 2π · [ ∫ sin³φ dφ − 2∫ cos²φ sin φ dφ ]
  = 2πa³ ( 2/3 − 2/3 ) = 0
```

**Answer: ∬_S (∇×F)·n dS = 0.**

## Check yourself

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

1. What does Stokes' theorem equate?
   A. Surface integral of curl F . n with the line integral of F . dr around the boundary
   B. Flux of F with volume integral of div F
   C. Line integral with area of C
   D. Gradient with curl
   Answer: A. Stokes relates curl flux through S with circulation around C. (Key point: Stokes' theorem)

2. What is the boundary of the upper hemisphere?
   A. The whole sphere
   B. Circle x^2 + y^2 = a^2 in the plane z = 0
   C. The point at the top
   D. The z-axis
   Answer: B. The open edge of the hemisphere is the equator circle. (Key point: Boundary)

3. On C with z = 0, what is F . dr?
   A. 2xy dt, giving pi a^2
   B. a^2 dt, giving 2pi a^2
   C. y dx + x dy = d(xy), which integrates to 0 round the closed curve
   D. x dy only
   Answer: C. An exact differential has zero integral over a closed loop. (Key point: Line integral)

4. What is curl F for F = y i + x(1-2z) j - xy k?
   A. 0
   B. y i + x j
   C. x i - y j + 2z k
   D. x i + y j - 2z k
   Answer: D. i: -x+2x = x; j: 0+y = y; k: (1-2z)-1 = -2z. (Key point: Curl)

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