---
title: "Double integral xy over region bounded by x²=4ay"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:double-integral-xy-over-region"
source: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-xy-over-region
site: BCT Engineering
---

# Double integral xy over region bounded by x²=4ay

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2072 Ashwin, 5 marks
- Video: [Evaluate ∬xydxdy where S is the domain bounded by the x-axis, ordinate x=2a, and the curve x²=4ay](https://www.youtube.com/watch?v=pAphMk3VgP4) — VisionPrep 📚
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-xy-over-region
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate: ∬_R xy dx dy where R is the region bounded by the x-axis, the ordinate x = 2a and the curve x² = 4ay.

## Short answer

Integrate over 0≤x≤2a, 0≤y≤x²/(4a). Inner integral yields x⁵/(32a²); outer integral gives a⁴/3.

## Answer

**Region.** Bounded by the x-axis (y = 0), the line x = 2a and the parabola x² = 4ay, i.e. y = x²/(4a). For each x in [0, 2a], y runs from 0 up to the parabola.

```text
I = ∫₀^{2a} ∫₀^{x²/4a} xy dy dx
```

#### Inner integral
```text
∫₀^{x²/4a} xy dy = x · (1/2)(x²/4a)² = x⁵/(32a²)
```

#### Outer integral
```text
I = (1/32a²) ∫₀^{2a} x⁵ dx
  = (1/32a²) · (2a)⁶/6
  = (1/32a²) · 64a⁶/6
  = a⁴/3
```

**Answer: a⁴/3.**

## Check yourself

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

1. y runs from 0 to:
   A. 2a
   B. x²/4a
   C. 4a/x²
   D. x²/a
   Answer: B. x² = 4ay. (Key point: Limits)

2. Inner integral:
   A. x³/(8a)
   B. x⁴/16a
   C. x⁵/(32a²)
   D. x⁵/(16a)
   Answer: C. x(1/2)(x²/4a)². (Key point: Inner)

3. Value:
   A. a⁶/3
   B. 2a⁴/3
   C. a⁴/6
   D. a⁴/3
   Answer: D. 64a⁶/(192a²). (Key point: Result)

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