---
title: "Area using Green's theorem"
subject: "Mathematics III"
chapter: 2
key: "semester_3/mathematics_iii:area-using-greens-theorem"
source: https://www.bctengineer.com/q/semester_3/mathematics_iii/area-using-greens-theorem
site: BCT Engineering
---

# Area using Green's theorem

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

- Asked in: 2082 Shrawan, 1+4 marks; 2081 Baishakh, 1+4 marks; 2080 Baishakh, 1+4 marks; 2078 Kartik, 5 marks; 2076 Asoj, 1+4 marks; 2072 Chaitra, 5 marks; 2072 Kartik, 5 marks; 2067 Asar (asked 8 times)
- Video: [Finding Areas with Green’s Theorem](https://www.youtube.com/watch?v=_aEgUhb6LtA) — Mathemation
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/area-using-greens-theorem
- Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-2

## Question

State Green's theorem and use it to find the area of the asteroid x^(2/3) + y^(2/3) = a^(2/3).

## Short answer

By Green’s theorem, area = ½ ∮ (x dy − y dx). For the asteroid x^(2/3)+y^(2/3)=a^(2/3), parameterize x = a cos^3 t, y = a sin^3 t, giving area = 3πa²/8 square units.

## Answer

#### Green's theorem
If C is a simple closed curve traversed counter-clockwise, enclosing a region R, and P, Q have continuous first partial derivatives in R, then

```text
∮_C (P dx + Q dy) = ∬_R (∂Q/∂x − ∂P/∂y) dx dy
```

#### Area formula
Take P = −y/2 and Q = x/2. Then ∂Q/∂x − ∂P/∂y = ½ + ½ = 1, so

```text
Area = ∬_R dx dy = ½ ∮_C (x dy − y dx)
```

#### Area of the astroid x^(2/3) + y^(2/3) = a^(2/3)
Parametric form: x = a cos³t, y = a sin³t, with t from 0 to 2π (this satisfies cos²t + sin²t = 1 after raising to 2/3).

```text
dx = −3a cos²t sin t dt
dy =  3a sin²t cos t dt
```

Then

```text
x dy − y dx = a cos³t · 3a sin²t cos t dt + a sin³t · 3a cos²t sin t dt
            = 3a² sin²t cos²t (cos²t + sin²t) dt
            = 3a² sin²t cos²t dt
            = (3a²/4) sin²2t dt
```

```text
Area = ½ ∫₀^{2π} (3a²/4) sin²2t dt
     = (3a²/8) ∫₀^{2π} (1 − cos 4t)/2 dt
     = (3a²/8) · [ t/2 − sin 4t/8 ]₀^{2π}
     = (3a²/8) · π
```

**Answer: Area of the astroid = 3πa²/8 square units.**

(Check: the astroid lies inside the circle of radius a, whose area is πa², and 3π/8 ≈ 1.18 < π ✓.)

## Check yourself

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

1. Choosing P=−y/2, Q=x/2 gives area =
   A. ∮(x dy + y dx)
   B. ½∮(x dy − y dx)
   C. ∮ x dy + y dx /2
   D. ½∮(y dy − x dx)
   Answer: B. ∂Q/∂x−∂P/∂y = 1. (Key point: Area formula)

2. The astroid x^(2/3)+y^(2/3)=a^(2/3) is parametrised as:
   A. x=a cos³t, y=a sin³t
   B. x=a cos t, y=a sin t
   C. x=a cos²t, y=a sin²t
   D. x=a³cos t, y=a³sin t
   Answer: A. Raising to 2/3 gives cos²t+sin²t=1. (Key point: Parametrisation)

3. x dy − y dx simplifies to:
   A. a² dt
   B. 3a² cos⁴t dt
   C. a² sin t cos t dt
   D. 3a² sin²t cos²t dt
   Answer: D. Using cos²t+sin²t=1. (Key point: Integrand)

4. The area is:
   A. πa²
   B. 3πa²/4
   C. 3πa²/8
   D. πa²/8
   Answer: C. ½·(3a²/4)·π·... = 3πa²/8. (Key point: Area)

## Related in Mathematics III

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## Related in other subjects

- [Parallel axis theorem; centroid of area between parabolas](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-centroid-curves-x-ky2) — Applied Mechanics, chapter 4
- [Parallel axis theorem; centroid of area under y = Kx²](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Aparallel-axis-centroid-y-kx2-ashad) — Applied Mechanics, chapter 4