---
title: "Area of astroid is (3/8)πa²"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:area-astroid-3-8-pi-a2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/area-astroid-3-8-pi-a2
site: BCT Engineering
---

# Area of astroid is (3/8)πa²

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2074 Asoj; 2067 Asoj (asked 2 times)
- Video: [Area of an Astroid: x^2/3+y^2/3=a^2/3](https://www.youtube.com/watch?v=dV2eBt3fprk) — Math Alchemist
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/area-astroid-3-8-pi-a2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Show that the area of the curve x^(2/3) + y^(2/3) = a^(2/3) is (3/8)πa².

## Short answer

Use x = a cos³t, y = a sin³t; area = 4∫₀^a y dx = 12a² ∫₀^{π/2} sin⁴t cos²t dt. Beta function yields π/32, giving area = 3πa²/8.

## Answer

Astroid: x^(2/3) + y^(2/3) = a^(2/3). It is symmetrical about both axes, so the total area is four times the area in the first quadrant.

Use x = a cos³t, y = a sin³t. Then dx = −3a cos²t sin t dt. As x goes from 0 to a, t goes from π/2 to 0.

```text
Area = 4 ∫₀^a y dx
     = 4 ∫_(π/2)^0 a sin³t (−3a cos²t sin t) dt
     = 12a² ∫₀^(π/2) sin⁴t cos²t dt
```
Beta function: ∫₀^(π/2) sin⁴t cos²t dt = Γ(5/2)Γ(3/2)/(2Γ(4)).

Γ(5/2) = 3√π/4, Γ(3/2) = √π/2, Γ(4) = 6.

= (3√π/4)(√π/2) / 12 = (3π/8)/12 = π/32.

Area = 12a² · π/32

**Area = 3πa²/8.** Shown.

## Check yourself

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

1. The total area of the astroid is
   A. 2 times the first quadrant area
   B. 8 times the first quadrant area
   C. 4 times the first quadrant area
   D. equal to the first quadrant area
   Answer: C. It is symmetric about both axes. (Key point: Symmetry)

2. Area = 4 integral y dx becomes
   A. 12a^2 integral sin^2 t cos^4 t dt
   B. 4a^2 integral sin^3 t cos^3 t dt
   C. 12a^2 integral sin^6 t dt
   D. 12a^2 integral sin^4 t cos^2 t dt over 0 to π/2
   Answer: D. y dx = a sin^3 t · 3a cos^2 t sin t dt. (Key point: Reduced integral)

3. integral sin^4 t cos^2 t dt over 0 to π/2 equals
   A. π/32
   B. π/16
   C. 3π/32
   D. π/8
   Answer: A. Γ(5/2)Γ(3/2)/(2Γ(4)) = (3π/8)/12 = π/32. (Key point: Beta value)

4. The area is
   A. 3πa^2/4
   B. 3πa^2/8
   C. πa^2/8
   D. 3πa^2/32
   Answer: B. 12a^2 · π/32 = 3πa^2/8. (Key point: Result)

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