Area of loop of ay²=x²(a−x)

Curve: ay² = x²(a − x), so y = ±(x/√a)√(a − x).

• It exists for x ≤ a. At x = 0 it has a node (double point at the origin) with tangents y² = x², i.e. y = ±x.
• It cuts the x-axis at x = 0 and x = a. For 0 ≤ x ≤ a, y is real, forming a loop between x = 0 and x = a, symmetrical about the x-axis.

Area of the loop = 2 ∫₀^a y dx = (2/√a) ∫₀^a x √(a − x) dx.

Put x = a sin²θ, dx = 2a sin θ cos θ dθ, √(a − x) = √a cos θ. Limits 0 to π/2.

∫₀^a x√(a−x) dx = ∫₀^(π/2) a sin²θ · √a cosθ · 2a sinθ cosθ dθ
                = 2a^(5/2) ∫₀^(π/2) sin³θ cos²θ dθ
                = 2a^(5/2) · Γ(2)Γ(3/2)/(2Γ(7/2))
                = 2a^(5/2) · (√π/2)/(2 · 15√π/8)
                = 2a^(5/2) · 2/15

Then Area = (2/√a) · (4/15) a^(5/2)

Area of the loop = 8a²/15 square units.