---
title: "Verify Stokes' theorem F=xi+z²j+y²k"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:verify-stokes-theorem-f-xi-z2j"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-stokes-theorem-f-xi-z2j
site: BCT Engineering
---

# Verify Stokes' theorem F=xi+z²j+y²k

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

- Asked in: 2067 Magh
- Video: [Stokes' Theorem Example // Verifying both Sides // Vector Calculus](https://www.youtube.com/watch?v=ms4JjH0BANU) — Dr. Trefor Bazett
- 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-xi-z2j
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Verify Stokes' theorem for F = x i + z² j + y² k over the plane surface x + y + z = 1 lying in the first octant.

## Short answer

curl F = (2y−2z)i. On plane n=(i+j+k)/√3, dS=√3 dx dy, (curl F)·n dS = (2y−2z)dxdy = (2x+4y−2)dxdy. Surface integral over triangle x,y≥0, x+y≤1 gives 0. Line integral over boundary is also 0. Stokes' theorem verified.

## Answer

**Stokes' theorem:** ∮_C F·dr = ∬_S (curl F)·n dS. Here F = x i + z² j + y² k and S is the triangle on x + y + z = 1 with vertices A(1,0,0), B(0,1,0), C(0,0,1).

#### Surface integral
```text
curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = (2y − 2z) i
n = (i + j + k)/√3,  dS = √3 dx dy
(curl F)·n dS = (2y − 2z) dx dy = (2y − 2 + 2x + 2y) dx dy = (2x + 4y − 2) dx dy
∬ over triangle x, y ≥ 0, x + y ≤ 1:
 = 2(1/6) + 4(1/6) − 2(1/2) = 1 − 1 = 0
```

#### Line integral (A → B → C → A)
F·dr = x dx + z² dy + y² dz.

```text
AB: z = 0, y = 1 − x, x: 1→0:   ∫ x dx = −1/2          (other terms 0)
BC: x = 0, y = 1 − z, z: 0→1:   ∫ z² dy + y² dz
                                = ∫ (−z²) dz + (1 − z)² dz = −1/3 + 1/3 = 0
CA: y = 0, z = 1 − x, x: 0→1:   ∫ x dx = 1/2           (other terms 0)
Total = −1/2 + 0 + 1/2 = 0
```

Both sides are 0, so **Stokes' theorem is verified.**

## Check yourself

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

1. Curl F for F = x i + z^2 j + y^2 k is:
   A. (2y + 2z) i
   B. 2z j
   C. (2y - 2z) i
   D. 0
   Answer: C. dR/dy - dQ/dz = 2y - 2z; other components zero. (Key point: Curl)

2. (curl F).n dS reduces to:
   A. sqrt(3)(2y-2z) dx dy
   B. (2y-2z)/sqrt3 dx dy
   C. (y - z) dx dy
   D. (2y - 2z) dx dy
   Answer: D. n has i-component 1/sqrt3 and dS = sqrt3 dx dy, so factors cancel. (Key point: Surface element)

3. On AB (z=0, y=1-x, x: 1 to 0) the integral is:
   A. -1/2
   B. 1/2
   C. 0
   D. -1
   Answer: A. Integral of x dx from 1 to 0. (Key point: Line integral AB)

4. Both sides are:
   A. 1
   B. 0
   C. 1/2
   D. -1/2
   Answer: B. Surface integral 0 and line total -1/2 + 0 + 1/2 = 0. (Key point: Conclusion)

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