---
title: "Verify Stokes' theorem F=(2x−y)i−yz²j−y²zk"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:verify-stokes-theorem-f-2x-y"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-stokes-theorem-f-2x-y
site: BCT Engineering
---

# Verify Stokes' theorem F=(2x−y)i−yz²j−y²zk

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

- Asked in: 2082 Shrawan, 5 marks; 2076 Asoj, 5 marks; 2073 Chaitra, 5 marks (asked 3 times)
- Video: [Verify Stoke's theorem for f=(2x-y)i - yz^2 j -  y^2 zk where S is upper half surface of the sphere](https://www.youtube.com/watch?v=MApB4d1QJpE) — Azhagappan Meyyappan
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-stokes-theorem-f-2x-y
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Verify Stokes' theorem for the vector field F = (2x − y) i − yz² j − y²z k, where S is the surface of the sphere x² + y² + z² = 1 above the xy-plane and C is its boundary.

## Short answer

The line integral around the circle x²+y²=1 with z=0 equals π. Curl F = k, so the surface integral over the projection disc also equals π, verifying Stokes’ theorem.

## Answer

**Stokes' theorem:** ∮_C F·dr = ∬_S (∇×F)·n dS, where C is the boundary of the open surface S, traversed counter-clockwise when seen from the tip of n.

F = (2x − y) i − yz² j − y²z k. S: upper hemisphere x² + y² + z² = 1, z ≥ 0. Boundary C: the circle x² + y² = 1, z = 0.

#### Left side: line integral around C
On C: x = cos t, y = sin t, z = 0, 0 ≤ t ≤ 2π. Then F·dr = (2x − y)dx − yz² dy − y²z dz = (2x − y)dx (the other terms vanish since z = 0, dz = 0).

```text
∮ F·dr = ∫₀^{2π} (2cos t − sin t)(−sin t) dt
       = ∫₀^{2π} (−2 sin t cos t + sin²t) dt
       = 0 + π = π
```

#### Right side: surface integral
```text
∇×F = | i       j       k    |
      | ∂/∂x   ∂/∂y   ∂/∂z  |
      | 2x−y   −yz²   −y²z  |
    = i(−2yz + 2yz) − j(0 − 0) + k(0 + 1) = k
```

So (∇×F)·n dS = n_z dS = dx dy (projection onto the xy-plane). The projection of the hemisphere is the unit disc, of area π:

```text
∬_S (∇×F)·n dS = ∬_{x²+y²≤1} dx dy = π
```

**Both sides equal π, so Stokes' theorem is verified.** ∎

## Check yourself

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

1. Stokes' theorem equates ∮F·dr with:
   A. ∭∇·F dV
   B. ∬(∇×F)·n dS
   C. ∬F·n dS
   D. ∬∇×F dV
   Answer: B. Circulation equals flux of the curl. (Key point: Statement)

2. Boundary of the upper hemisphere is:
   A. Circle x²+y²=1, z=0
   B. Sphere itself
   C. The z-axis
   D. None
   Answer: A. Hemisphere z≥0 meets the xy-plane in unit circle. (Key point: Boundary)

3. ∇×F for F=(2x−y)i−yz²j−y²zk:
   A. 0
   B. 2k
   C. i+j
   D. k
   Answer: D. Only the k component −(−1)=1 survives. (Key point: Curl)

4. Both sides equal:
   A. 2π
   B. 0
   C. π
   D. π/2
   Answer: C. Line integral gives π and surface integral gives disc area π. (Key point: Both sides)

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