Find the centroid (x̄, ȳ) of the region x² + y² ≤ 1 with x ≥ 0, y ≥ 0.
The quarter disk is bounded by x² + y² ≤ 1 in the first quadrant: 0 ≤ x, y.
x̄ = (1/A) ∬R x dA and ȳ = (1/A) ∬R y dA
- x = r cosθ, y = r sinθ, dA = r dr dθ
- Bounds: 0 ≤ r ≤ 1, 0 ≤ θ ≤ π/2
A = ∬ dA = ∫0π/2 ∫01 r dr dθ = ∫0π/2 [ r²/2 ]01 dθ = ∫0π/2 (1/2) dθ = π/4
x̄ = (1/A) ∬ x dA = (1/A) ∫0π/2 ∫01 (r cosθ) · r dr dθ = (1/A) ∫0π/2 cosθ ∫01 r² dr dθ = (1/A) ∫0π/2 cosθ · (1/3) dθ = (1/(3A)) [ sinθ ]0π/2 = 1/(3A) = 1 / (3·π/4) = 4/(3π)
By symmetry of the quarter circle in the first quadrant, ȳ = x̄.
ȳ = 4/(3π)
Centroid: (x̄, ȳ) = (4/(3π), 4/(3π))
