---
title: "Area bounded by x²y=a²(a−y) and x-axis"
subject: "Mathematics I"
chapter: 2
key: "semester_1/mathematics_i:area-x2y-a2-a-minus-y"
source: https://www.bctengineer.com/q/semester_1/mathematics_i/area-x2y-a2-a-minus-y
site: BCT Engineering
---

# Area bounded by x²y=a²(a−y) and x-axis

Mathematics I · Chapter 2: Integration and its Applications

- Asked in: 2070 Asar; 2069 Chaitra (asked 2 times)
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_1/mathematics_i/area-x2y-a2-a-minus-y
- Whole chapter: https://www.bctengineer.com/md/semester_1/mathematics_i/chapter-2

## Question

Find the area bounded by the curve x²y = a²(a−y) and the X-axis.

## Short answer

Rewrite as y = a³/(x²+a²). The area is ∫_{-∞}^∞ a³/(x²+a²) dx = 2a³ ∫₀^∞ dx/(x²+a²) = πa².

## Answer

Curve: x²y = a²(a − y). Solve for y:

x²y + a²y = a³, so **y = a³/(x² + a²)** (the Witch of Agnesi).

The curve lies above the x-axis (y > 0), is symmetric about the y-axis, and y → 0 as x → ±∞; so the x-axis is an asymptote. The area between the curve and the x-axis is

```text
A = ∫_(−∞)^(∞) a³/(x² + a²) dx
  = 2a³ ∫₀^∞ dx/(x² + a²)
  = 2a³ · (1/a)[tan⁻¹(x/a)]₀^∞
  = 2a² · (π/2)
```
**Area = πa² square units.**

## Check yourself

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

1. Solving x^2 y = a^2(a - y) gives
   A. y = a^2/(x^2+a^2)
   B. y = a^3/(x^2-a^2)
   C. y = a^3/(x^2+a^2)
   D. y = a/(x^2+a^2)
   Answer: C. Collect y terms: y(x^2 + a^2) = a^3. (Key point: Solve for y)

2. The x-axis is an asymptote because
   A. y = 0 at x = 0
   B. the curve is symmetric about the y-axis
   C. y is negative for large x
   D. y → 0 as x → ±∞
   Answer: D. The curve approaches the axis without touching. (Key point: Asymptote)

3. The area is taken over
   A. x from -∞ to ∞
   B. x from 0 to a
   C. x from -a to a
   D. x from 0 to 1
   Answer: A. The region extends infinitely along the axis. (Key point: Limits)

4. The area is
   A. πa^2/2
   B. πa^2
   C. 2πa^2
   D. πa^3
   Answer: B. 2a^3 · (1/a)(π/2) = πa^2. (Key point: Result)

## Related in Mathematics I

- [Area between x²=4y and x=4y−2](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aarea-parabola-and-line) — Mathematics I, chapter 2
- [Curvature where tangent is parallel to x-axis](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Acurvature-y-x2-x-minus-3) — Mathematics I, chapter 1
- [Area of astroid is (3/8)πa²](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aarea-astroid-3-8-pi-a2) — Mathematics I, chapter 2, asked 2×
- [Area between y²(2a−x)=x³ and its asymptote](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aarea-asymptote-curve-y2-2a-minus-x) — Mathematics I, chapter 2
- [Area of one loop of r = a sin 3θ](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aarea-one-loop-sin-3theta) — Mathematics I, chapter 2
- [Area of two loops of a²y²=a²x²−x⁴](https://www.bctengineer.com/question.md?key=semester_1%2Fmathematics_i%3Aarea-two-loops-a2y2) — Mathematics I, chapter 2

## Related in other subjects

- [Area bounded by y²=x³ and y=x](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Aarea-bounded-y2-x3-y) — Mathematics, chapter 2
- [Smaller area bounded by circle and line](https://www.bctengineer.com/question.md?key=semester_2%2Fmathematics%3Asmaller-area-bounded-circle-line) — Mathematics, chapter 2
- [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
- [MOI of composite area about Y axis](https://www.bctengineer.com/question.md?key=semester_1%2Fapplied_mechanics%3Amoi-composite-area-about-y-axis) — Applied Mechanics, chapter 4