---
title: "Area of loop of ay²=x²(a−x)"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:area-loop-ay2-x2-a-minus-x"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/area-loop-ay2-x2-a-minus-x
site: BCT Engineering
---

# Area of loop of ay²=x²(a−x)

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2067 Asar
- Video: [Find the area of the loop of the loop of the curve ay² = x²(a -x ).](https://www.youtube.com/watch?v=l6KyYpuBs_Q) — Mathematics Zone by keshri sir
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/area-loop-ay2-x2-a-minus-x
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find, by method of integration, the area of the loop of the curve ay² = x²(a−x).

## Short answer

The loop lies in 0≤x≤a. Area = 2∫₀^a y dx = (2/√a)∫₀^a x√(a-x) dx; substitute x = a sin²θ to obtain 8a²/15 square units.

## Answer

Curve: ay² = x²(a − x), so y = ±(x/√a)√(a − x).

- It exists for x ≤ a. At x = 0 it has a node (double point at the origin) with tangents y² = x², i.e. y = ±x.
- It cuts the x-axis at x = 0 and x = a. For 0 ≤ x ≤ a, y is real, forming a loop between x = 0 and x = a, symmetrical about the x-axis.

Area of the loop = 2 ∫₀^a y dx = (2/√a) ∫₀^a x √(a − x) dx.

Put x = a sin²θ, dx = 2a sin θ cos θ dθ, √(a − x) = √a cos θ. Limits 0 to π/2.

```text
∫₀^a x√(a−x) dx = ∫₀^(π/2) a sin²θ · √a cosθ · 2a sinθ cosθ dθ
                = 2a^(5/2) ∫₀^(π/2) sin³θ cos²θ dθ
                = 2a^(5/2) · Γ(2)Γ(3/2)/(2Γ(7/2))
                = 2a^(5/2) · (√π/2)/(2 · 15√π/8)
                = 2a^(5/2) · 2/15
```
Then Area = (2/√a) · (4/15) a^(5/2)

**Area of the loop = 8a²/15 square units.**

## Check yourself

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

1. At the origin the curve has
   A. a cusp
   B. a node with tangents y = ±x
   C. an inflection
   D. an isolated point
   Answer: B. Lowest terms y^2 = x^2. (Key point: Node)

2. Loop area equals
   A. integral y dx from -a to a
   B. (1/sqrt a) integral x^2 dx
   C. (2/sqrt a) integral x sqrt(a-x) dx from 0 to a
   D. 2 integral from 0 to ∞
   Answer: C. Double the upper half. (Key point: Setup)

3. x = a sin^2 θ gives
   A. a^(5/2) integral sin^2 θ cos θ dθ
   B. 2a^2 integral sin^3 θ dθ
   C. 2a^(5/2) integral sin^2 θ cos^3 θ dθ
   D. 2a^(5/2) integral sin^3 θ cos^2 θ dθ
   Answer: D. x√(a-x)dx. (Key point: Substitution)

4. Loop area is
   A. 8a^2/15
   B. 4a^2/15
   C. 16a^2/15
   D. 2a^2/3
   Answer: A. (2/√a)(4/15)a^(5/2). (Key point: Result)

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