---
title: "Apply Green's theorem F=(x²−xy³)i+(y²−2xy)j"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:apply-greens-theorem-f-x2-xy3"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/apply-greens-theorem-f-x2-xy3
site: BCT Engineering
---

# Apply Green's theorem F=(x²−xy³)i+(y²−2xy)j

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

- Asked in: 2079 Bhadra, 5 marks
- Video: [Greens theorem problem||Vector Integration || Rajeshlekkalu](https://www.youtube.com/watch?v=9W9QC-GqoKs) — Rajesh Lekkalu
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/apply-greens-theorem-f-x2-xy3
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

Apply Green's theorem in plane to evaluate ∮_C F·dr, where F = (x² − xy³) i + (y² − 2xy) j and C is a square with vertices (0,0), (2,0), (2,2), (0,2).

## Short answer

Apply Green's theorem: ∂Q/∂x − ∂P/∂y = −2y + 3xy². The double integral over the square 0≤x,y≤2 equals 8, so ∮_C F·dr = 8.

## Answer

**Green's theorem:** ∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy.

Here F·dr = (x² − xy³) dx + (y² − 2xy) dy.

```text
P = x² − xy³   → ∂P/∂y = −3xy²
Q = y² − 2xy   → ∂Q/∂x = −2y
∂Q/∂x − ∂P/∂y = −2y + 3xy²
```

Region R: the square 0 ≤ x ≤ 2, 0 ≤ y ≤ 2.

```text
∬_R (−2y + 3xy²) dx dy = ∫₀² ∫₀² (−2y + 3xy²) dx dy
inner (x): [−2xy + 3x²y²/2]₀² = −4y + 6y²
outer (y): ∫₀² (−4y + 6y²) dy = [−2y² + 2y³]₀² = −8 + 16 = 8
```

**Answer: ∮_C F·dr = 8.**

## Check yourself

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

1. For Q = y^2 - 2xy, dQ/dx?
   A. -2y
   B. 2y
   C. -2x
   D. y
   Answer: A. Differentiate in x. (Key point: dQ/dx)

2. For P = x^2 - xy^3, dP/dy?
   A. 3xy^2
   B. -3xy^2
   C. -xy^2
   D. -3y^2
   Answer: B. Differentiate in y. (Key point: dP/dy)

3. dQ/dx - dP/dy?
   A. -2y - 3xy^2
   B. 2y + 3xy^2
   C. -2y + 3xy^2
   D. -2y
   Answer: C. Subtract. (Key point: Integrand)

4. Value?
   A. 0
   B. 16
   C. -8
   D. 8
   Answer: D. Outer integral gives -8 + 16. (Key point: Result)

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