---
title: "Verify Stokes' theorem F=(x²+y²)i−2xyj"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:verify-stokes-theorem-f-x2-y2"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/verify-stokes-theorem-f-x2-y2
site: BCT Engineering
---

# Verify Stokes' theorem F=(x²+y²)i−2xyj

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

- Asked in: 2079 Bhadra, 5 marks; 2074 Chaitra, 5 marks; 2071 Chaitra, 5 marks (asked 3 times)
- Video: [Verify Stokes theorem for F = (x²+y²)i - 2xy j  | Surface Integrals | Vector Calculus](https://www.youtube.com/watch?v=m3i-CYczr1s) — Spectrum of Mathematics
- 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-x2-y2
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Verify Stokes' theorem for F = (x² + y²) i − 2xy j taken round the rectangle bounded by the lines x = ±a, y = 0, y = b.

## Short answer

Line integration along the rectangle edges yields −4ab². Curl F = −4y k, whose double integral over the rectangle gives −4ab², confirming the theorem.

## Answer

**Stokes' theorem (plane form):** ∮_C F·dr = ∬_S (∇×F)·k dx dy, where C bounds S in the xy-plane.

F = (x² + y²) i − 2xy j. The rectangle has vertices A(−a, 0), B(a, 0), C(a, b), D(−a, b).

#### Left side: line integral (counter-clockwise)
F·dr = (x² + y²) dx − 2xy dy.

```text
AB : y = 0, x from −a to a, dy = 0
     ∫ x² dx = [x³/3]₋ₐᵃ = 2a³/3

BC : x = a, y from 0 to b, dx = 0
     ∫ −2a y dy = −a b²

CD : y = b, x from a to −a, dy = 0
     ∫ (x² + b²) dx = [x³/3 + b²x]ₐ^{−a} = −(2a³/3 + 2ab²)

DA : x = −a, y from b to 0, dx = 0
     ∫ −2(−a) y dy = ∫_b^0 2a y dy = −a b²
```

```text
Total = 2a³/3 − ab² − 2a³/3 − 2ab² − ab² = −4ab²
```

#### Right side: surface integral
```text
∇×F = k ( ∂Q/∂x − ∂P/∂y ) = k ( −2y − 2y ) = −4y k

∬ (∇×F)·k dx dy = ∫₀^b ∫₋ₐᵃ (−4y) dx dy
                = −4 · 2a · [y²/2]₀^b = −4 · 2a · b²/2 = −4ab²
```

**Both sides equal −4ab², so Stokes' theorem is verified.** ∎

## Check yourself

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

1. For a plane region, ∮F·dr equals:
   A. ∬(∂P/∂x+∂Q/∂y) dx dy
   B. ∬(∂Q/∂x − ∂P/∂y) dx dy
   C. ∮ P dx only
   D. 0
   Answer: B. This is Green's theorem in plane form. (Key point: Plane Stokes)

2. ∇×F·k for F=(x²+y²)i−2xy j:
   A. −4y
   B. −2y
   C. 4y
   D. 0
   Answer: A. −2y − 2y = −4y. (Key point: Curl)

3. On BC (y=b, x from a to −a) integral is:
   A. 2a³/3+2ab²
   B. 0
   C. −2ab²
   D. −(2a³/3+2ab²)
   Answer: D. Direction is from a to −a. (Key point: Side BC)

4. Both sides equal:
   A. 4ab²
   B. −2ab²
   C. −4ab²
   D. 0
   Answer: C. The line integral and surface integral agree. (Key point: Total)

## Close variants set in other exams

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