---
title: "Double integral xy over ellipse in first quadrant"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:double-integral-xy-over-ellipse"
source: https://www.bctengineer.com/q/semester_2/mathematics/double-integral-xy-over-ellipse
site: BCT Engineering
---

# Double integral xy over ellipse in first quadrant

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2070 Magh, 5 marks; 2067 Mangsir, 5 marks; 2067 Chaitra, 5 marks (asked 3 times)
- Video: [VTU Engineering Maths 1 Evaluate double integration of  xy over the positive quadrant of the circle](https://www.youtube.com/watch?v=UWIL3sf-Ups) — EASY MATHS EASY TRICKS
- 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-ellipse
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate: ∬_R xy dx dy where R is the region over the area of the ellipse x²/a² + y²/b² = 1 in the first quadrant.

## Short answer

Region: first quadrant inside x²/a² + y²/b² = 1. Integrate xy dy dx: inner integral yields (x b²/2)(1−x²/a²), outer gives a²b²/8.

## Answer

**Region.** The first quadrant part of the ellipse x²/a² + y²/b² = 1: x from 0 to a, and y from 0 to y_max = (b/a)√(a² − x²) = b√(1 − x²/a²).

```text
I = ∫₀^a ∫₀^{b√(1−x²/a²)} xy dy dx
```

#### Inner integral
```text
∫₀^{y_max} xy dy = x · y_max²/2 = (x b²/2)(1 − x²/a²)
```

#### Outer integral
```text
I = (b²/2) ∫₀^a ( x − x³/a² ) dx
  = (b²/2) [ x²/2 − x⁴/(4a²) ]₀^a
  = (b²/2) ( a²/2 − a²/4 )
  = (b²/2)(a²/4)
  = a²b²/8
```

**Answer: a²b²/8.**

## Check yourself

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

1. In the first quadrant, y runs from 0 to:
   A. a√(1 − x²/b²)
   B. b√(1 − x²/a²)
   C. b(1 − x²/a²)
   D. b
   Answer: B. Solve the ellipse for y. (Key point: Limits)

2. ∫ xy dy from 0 to y_max equals:
   A. x y_max
   B. x² y_max/2
   C. x y_max²/2
   D. y_max²/2
   Answer: C. x is constant. (Key point: Inner integral)

3. The value is:
   A. a²b²/8
   B. a²b²/2
   C. ab/8
   D. a²b²/4
   Answer: A. (b²/2)(a²/2 − a²/4) = a²b²/8. (Key point: Result)

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