Polar coordinates: √(x²+y²) integral

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.

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

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.

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

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.

✓ Answer checked by Claude Opus 5.5 (high effort)

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