---
title: "Evaluate using Stokes' theorem"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:evaluate-using-stokes-theorem"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/evaluate-using-stokes-theorem
site: BCT Engineering
---

# Evaluate using Stokes' theorem

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

- Asked in: 2080 Bhadra, 5 marks
- Video: [Stoke's Theorem Problem 1 |  Vector Claculus | Engineering Mathematics](https://www.youtube.com/watch?v=DRWfHYiXnBs) — Alex Maths Engineering
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/evaluate-using-stokes-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Evaluate ∮_C (xy dx + xy² dy) by Stokes' theorem, where C is the square in the xy-plane with vertices (1,1), (−1,1), (−1,−1) and (1,−1).

## Short answer

Curl F = (y^2 − x) k. Surface integral over the square (−1≤x,y≤1) gives ∬(y^2 − x) dxdy = 4/3.

## Answer

**Stokes' theorem:** ∮_C F·dr = ∬_S (curl F)·n dS. Write the integral as ∮_C (xy dx + xy² dy + 0 dz), so F = xy i + xy² j.

```text
curl F = (∂R/∂y − ∂Q/∂z) i + (∂P/∂z − ∂R/∂x) j + (∂Q/∂x − ∂P/∂y) k
       = 0 i + 0 j + (y² − x) k
```

C is the square with vertices (1,1), (−1,1), (−1,−1), (1,−1), traversed anticlockwise. Take S as the flat square region in the xy-plane, so n = k and dS = dx dy.

```text
∮_C = ∬_S (y² − x) dx dy
    = ∫₋₁¹ ∫₋₁¹ (y² − x) dx dy
    = ∫₋₁¹ [y² x − x²/2]₋₁¹ dy
    = ∫₋₁¹ 2y² dy
    = 2 · (2/3)
    = 4/3
```

**Answer: ∮_C (xy dx + xy² dy) = 4/3.**

## Check yourself

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

1. From the integral of xy dx + xy^2 dy, F is?
   A. xy i + xy^2 j
   B. xy^2 i + xy j
   C. x i + y j
   D. xy i + y^2 j
   Answer: A. Coefficients of dx and dy. (Key point: Reading F)

2. curl F?
   A. (y^2 + x) k
   B. (y^2 - x) k
   C. (x - y^2) k
   D. y^2 k
   Answer: B. dQ/dx - dP/dy = y^2 - x. (Key point: Curl)

3. Why n = k?
   A. The curve is a circle
   B. The square is vertical
   C. S is the flat square region in the xy-plane
   D. F is along k
   Answer: C. Flat region in xy-plane. (Key point: Surface choice)

4. Value?
   A. 0
   B. 2/3
   C. 8/3
   D. 4/3
   Answer: D. x term vanishes, 2y^2 integrates over y to 4/3. (Key point: Result)

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