---
title: "Area of two loops of a²y²=a²x²−x⁴"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:area-two-loops-a2y2"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/area-two-loops-a2y2
site: BCT Engineering
---

# Area of two loops of a²y²=a²x²−x⁴

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2070 Chaitra
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/area-two-loops-a2y2
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the area of two loops of the curve a²y² = a²x² − x⁴.

## Short answer

By symmetry, total area = 4∫₀^a y dx = (4/a)∫₀^a x√(a²-x²) dx. Substituting t=a²-x² yields 4a²/3 square units.

## Answer

Curve: a²y² = a²x² − x⁴ = x²(a² − x²). So y = ±(x/a)√(a² − x²).

The curve exists only for −a ≤ x ≤ a. It is symmetrical about both axes and passes through the origin (a node), forming two loops, one on each side of the y-axis (the figure-of-eight shape).

By symmetry, the total area of the two loops is four times the area in the first quadrant:

```text
Area = 4 ∫₀^a y dx
     = (4/a) ∫₀^a x √(a² − x²) dx
```
Put a² − x² = t, −2x dx = dt. Limits: x = 0 → t = a²; x = a → t = 0.

```text
∫₀^a x√(a²−x²) dx = ½ ∫₀^(a²) t^(1/2) dt = ½ · (2/3) a³ = a³/3
```
Area = (4/a)(a³/3)

**Area of the two loops = 4a²/3 square units** (each loop has area 2a²/3).

## Check yourself

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

1. The curve exists only for
   A. all x
   B. -a ≤ x ≤ a
   C. x ≥ 0
   D. |x| ≥ a
   Answer: B. a^2 - x^2 must be non-negative. (Key point: Domain)

2. The total area equals
   A. 2 times
   B. 8 times
   C. 4 times the first quadrant area
   D. the first quadrant area
   Answer: C. Symmetric about both axes. (Key point: Symmetry)

3. Put a^2 - x^2 = t; the integral of x sqrt(a^2-x^2) becomes
   A. integral t^(1/2) dt
   B. (1/2) integral t^(-1/2) dt
   C. 2 integral t^(1/2) dt
   D. (1/2) integral t^(1/2) dt
   Answer: D. -2x dx = dt. (Key point: Substitution)

4. The area is
   A. 4a^2/3
   B. 2a^2/3
   C. 8a^2/3
   D. a^2/3
   Answer: A. (4/a)(a^3/3). (Key point: Result)

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