---
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-two"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-stokes-theorem-f-2x-y-two
site: BCT Engineering
---

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

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

- Asked in: 2070 Chaitra, 5 marks; 2067 Asar (asked 2 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-two
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Verify Stokes' theorem for F = (2x − y) i − yz² j − y²z k where S is the upper part of the sphere x² + y² + z² = a² and C is its boundary.

## Short answer

Line integral around boundary circle gives πa². Curl F = k, so surface flux over the disc projection equals πa². Both sides match πa², verifying Stokes' theorem.

## Answer

**Stokes' theorem:** ∮_C F·dr = ∬_S (∇×F)·n dS.

F = (2x − y) i − yz² j − y²z k, S: upper hemisphere x² + y² + z² = a², z ≥ 0. Boundary C: the circle x² + y² = a², z = 0.

#### Left side: line integral
On C, z = 0 and dz = 0, so F·dr = (2x − y) dx. Put x = a cos t, y = a sin t, 0 ≤ t ≤ 2π, dx = −a sin t dt:

```text
∮ F·dr = ∫₀^{2π} (2a cos t − a sin t)(−a sin t) dt
       = a² ∫₀^{2π} (−2 sin t cos t + sin²t) dt
       = a² (0 + π) = πa²
```

#### Right side: surface integral
```text
∇×F = i(∂R/∂y − ∂Q/∂z) + j(∂P/∂z − ∂R/∂x) + k(∂Q/∂x − ∂P/∂y)
    = 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 disc x² + y² ≤ a²:

```text
∬_S (∇×F)·n dS = ∬_{x²+y²≤a²} dx dy = πa²
```

**Both sides equal πa², so Stokes' theorem is verified.** ∎

## Check yourself

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

1. On C (z = 0), what does F.dr reduce to?
   A. (2x-y) dx - yz^2 dy
   B. 2x dx + y dy
   C. 0
   D. (2x - y) dx
   Answer: D. With z = 0 and dz = 0, only P dx remains in the plane integrand (the j term has factor z^2). (Key point: Boundary line integral)

2. What is curl F?
   A. k
   B. 0
   C. i + j + k
   D. -k
   Answer: A. The i term -2yz + 2yz = 0, j term 0, k term 0 + 1 = 1. (Key point: Curl)

3. Why does the surface integral equal the area of a disc?
   A. The hemisphere is flat
   B. curl F . n dS = n_z dS = dx dy, the projection onto the xy-plane
   C. curl F is zero
   D. dS = a dtheta
   Answer: B. Projecting onto the xy-plane gives the disc of radius a. (Key point: Surface integral)

4. Both sides equal what?
   A. 2 pi a^2
   B. pi a
   C. pi a^2
   D. 0
   Answer: C. Line integral gives a^2 times pi and surface integral gives the disc area. (Key point: Result)

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