---
title: "Change order: y²/√(y⁴−a²x²)"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:change-order-y2-sqrt-y"
source: https://www.bctengineer.com/q/semester_2/mathematics/change-order-y2-sqrt-y
site: BCT Engineering
---

# Change order: y²/√(y⁴−a²x²)

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2075 Baisakh, 5 marks; 2070 Bhadra, 5 marks (asked 2 times)
- Video: [Change of order of Integration in Hindi | double integral | Multiple integral | part 17|Barun maths](https://www.youtube.com/watch?v=QlM-m2bYUAM) — Dr Barun Kumar yadav
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/change-order-y2-sqrt-y
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate ∫∫ y²/√(y⁴−a²x²) dy dx by changing order of integration. [Limits as printed: outer integral from 0 to a, inner from √(ax) to a; scan unclear]

## Short answer

Limits: 0≤x≤a, √(ax)≤y≤a. Reverse order: 0≤y≤a, 0≤x≤y²/a. Inner integral ∫ dx/√(y⁴−a²x²) = π/(2a). Then ∫₀^a y²*(π/(2a)) dy = π a²/6.

## Answer

**How I read the limits.** As printed, the integral is I = ∫₀^a ∫_{√(ax)}^{a} y²/√(y⁴ − a²x²) dy dx. This reading is consistent: the integrand is real exactly when y⁴ ≥ a²x², i.e. y ≥ √(ax), which is the lower limit of y.

#### Region
0 ≤ x ≤ a and √(ax) ≤ y ≤ a. The lower boundary y = √(ax) is the parabola y² = ax, so x ≤ y²/a. Taking horizontal strips: y from 0 to a, and x from 0 to y²/a.

```text
y
a |- - - - - -*(a,a)
  |         . '
  |     . '   region: between x=0
  |  . '      and parabola y² = ax
  |_'_ _ _ _ _ _ x
 0               a
```

#### Reversed order
```text
I = ∫₀^a y² [ ∫₀^{y²/a} dx / √(y⁴ − a²x²) ] dy
```

Inner: ∫ dx/√(y⁴ − a²x²) = (1/a) sin⁻¹(ax/y²). Between 0 and y²/a: (1/a)(sin⁻¹ 1 − 0) = π/(2a).

```text
I = ∫₀^a y² · π/(2a) dy
  = (π/2a)(a³/3)
  = π a²/6
```

**Answer: π a²/6.**

## Check yourself

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

1. The region 0≤x≤a, √(ax)≤y≤a is bounded by:
   A. The line y = ax
   B. y = √(ax), x=0 and y=a
   C. A circle about the origin
   D. The line x = a only
   Answer: B. Parabola y² = ax. (Key point: Region)

2. After reversal x runs:
   A. y²/a to a
   B. 0 to y²/a with y from 0 to a
   C. 0 to a with y from 0 to a
   D. 0 to √(ay)
   Answer: B. x ≤ y²/a. (Key point: Reversed limits)

3. ∫₀^{y²/a} dx/√(y⁴−a²x²) equals:
   A. 1/a
   B. π/a
   C. π/(2a)
   D. π/2
   Answer: C. (1/a) sin⁻¹(1). (Key point: Inner integral)

4. The value is:
   A. πa³/6
   B. πa²/3
   C. πa/6
   D. πa²/6
   Answer: D. ∫ y² π/(2a) dy = πa²/6. (Key point: Result)

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