---
title: "Work done by a force field F=(2x−y+z)i+(x+y−z²)j…"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:work-done-by-a-force-field"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/work-done-by-a-force-field
site: BCT Engineering
---

# Work done by a force field F=(2x−y+z)i+(x+y−z²)j…

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

- Asked in: 2070 Asar, 5 marks; 2067 Asar (asked 2 times)
- Video: [Line Integral Solved Problem: Finding Work Done Along a Straight Line Path by team education](https://www.youtube.com/watch?v=uRkW9kqwa-A) — 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/work-done-by-a-force-field
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Find the work done in moving a particle once round the circle x² + y² = 9, z = 0 under the force field F given by F = (2x − y + z) i + (x + y − z²) j + (3x − 2y + 4z) k.

## Short answer

By Stokes' theorem, curl F·k = 2. The disc area is 9π, so work = 18π (anticlockwise). Direct line integral confirms 18π.

## Answer

Work done = ∮_C F·dr around the closed circle. Use Stokes' theorem to avoid parametrising the three components, then check directly.

**By Stokes' theorem** (C counter-clockwise, S = disc x² + y² ≤ 9 in the plane z = 0, n = k):

```text
F = (2x − y + z) i + (x + y − z²) j + (3x − 2y + 4z) k
curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (−2 + 2z) i + (1 − 3) j + (1 + 1) k
       = (2z − 2) i − 2 j + 2 k
curl F · k = 2
Work = ∬_S (curl F)·k dA = 2 × (area of disc) = 2 × π(3)² = 18π
```

**Direct check.** On the circle z = 0: x = 3cos t, y = 3sin t, 0 ≤ t ≤ 2π, dz = 0.

```text
F·dr = (2x − y) dx + (x + y) dy
     = (6cos t − 3sin t)(−3sin t) dt + (3cos t + 3sin t)(3cos t) dt
     = (−18 sin t cos t + 9 sin²t + 9 cos²t + 9 sin t cos t) dt
     = (9 − 9 sin t cos t) dt
∫₀^2π (9 − 9 sin t cos t) dt = 18π − 0 = 18π
```

**Answer: Work done = 18π ≈ 56.55 units** (anticlockwise direction).

## Check yourself

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

1. Work done around a closed curve is?
   A. The divergence
   B. div F times area
   C. F times length
   D. The line integral of F.dr around the curve
   Answer: D. Work is the line integral of force along the path. (Key point: Work)

2. What is (curl F).k for this F?
   A. 2
   B. -2
   C. 1
   D. 3
   Answer: A. dQ/dx - dP/dy = 1 - (-1) = 2. (Key point: Curl k-component)

3. Using Stokes, the work equals?
   A. 2 x 3 pi
   B. 2 times the area of the disc, 2 x 9pi = 18pi
   C. 9pi
   D. 2 x 6pi = 12pi
   Answer: B. Disc radius 3 has area 9pi. (Key point: Stokes use)

4. On z = 0, F.dr becomes?
   A. 9 sin t cos t dt
   B. 0
   C. (9 - 9 sin t cos t) dt
   D. 18 dt sin t
   Answer: C. The sin t cos t terms integrate to 0 and 9 integrates to 18pi. (Key point: Direct check)

## Related in Mathematics III

- [Work done by a force field F=(2x−y+2z)i+(x+y−z)j…](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Awork-done-by-a-force-field-two) — Mathematics III, chapter 2
- [Work done by a force field F=yzi+zxj+xyk](https://www.bctengineer.com/question.md?key=semester_3%2Fmathematics_iii%3Awork-done-by-a-force-field-three) — Mathematics III, chapter 2

## Related in other subjects

- [Jet on moving curved vanes, force and work](https://www.bctengineer.com/question.md?key=semester_3%2Ffluid_mechanics%3Ajet-on-moving-curved-vanes) — Fluid Mechanics, chapter 7
- [Velocity and acceleration of moving particle at t=1](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Avelocity-acceleration-moving-particle-t) — Mathematics, chapter 5