---
title: "Polar coordinates: √(x²+y²) integral"
subject: "Mathematics"
chapter: 2
key: "semester_2/mathematics:polar-coordinates-sqrt-x2-y2"
source: https://www.bctengineer.com/q/semester_2/mathematics/polar-coordinates-sqrt-x2-y2
site: BCT Engineering
---

# Polar coordinates: √(x²+y²) integral

Mathematics · Chapter 2: Multiple Integrals

- Asked in: 2073 Magh, 5 marks; 2071 Magh, 5 marks (asked 2 times)
- Video: [How to evaluate a double integral using polar coordinates](https://www.youtube.com/watch?v=CMTnumu1n68) — bprp calculus basics
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_2/mathematics/polar-coordinates-sqrt-x2-y2
- Whole chapter: https://www.bctengineer.com/md/semester_2/mathematics/chapter-2

## Question

Evaluate the integral by changing to polar co-ordinates: ∫₀ᵃ ∫₀^√(a²−x²) √(x²+y²) dy dx.

## Short answer

Quarter disc in first quadrant: 0≤r≤a, 0≤θ≤π/2. Integrand √(x²+y²)=r, area element r dr dθ. ∫₀^{π/2}∫₀^a r² dr dθ = π a³/6.

## Answer

**Region.** x from 0 to a, y from 0 to √(a² − x²): the quarter disc x² + y² ≤ a² in the first quadrant. Polar limits: 0 ≤ r ≤ a, 0 ≤ θ ≤ π/2.

#### Change of variables
x = r cos θ, y = r sin θ, √(x² + y²) = r, dy dx = r dr dθ.

```text
I = ∫₀^{π/2} ∫₀^a r · r dr dθ
  = ∫₀^{π/2} ∫₀^a r² dr dθ
  = ∫₀^{π/2} dθ · [ r³/3 ]₀^a
  = (π/2)(a³/3)
  = π a³/6
```

**Answer: π a³/6.**

## Check yourself

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

1. dy dx becomes:
   A. dr dθ
   B. r dr dθ
   C. r² dr dθ
   D. dr dθ / r
   Answer: B. Jacobian r. (Key point: Element)

2. The polar integrand is:
   A. r³
   B. r
   C. 1
   D. r²
   Answer: D. r · r. (Key point: Integrand)

3. The value is:
   A. πa³/3
   B. a³/3
   C. πa²/4
   D. πa³/6
   Answer: D. (π/2)(a³/3). (Key point: Result)

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