Area of two loops of a²y²=a²x²−x⁴

Curve: a²y² = a²x² − x⁴ = x²(a² − x²). So y = ±(x/a)√(a² − x²).

The curve exists only for −a ≤ x ≤ a. It is symmetrical about both axes and passes through the origin (a node), forming two loops, one on each side of the y-axis (the figure-of-eight shape).

By symmetry, the total area of the two loops is four times the area in the first quadrant:

Area = 4 ∫₀^a y dx
     = (4/a) ∫₀^a x √(a² − x²) dx

Put a² − x² = t, −2x dx = dt. Limits: x = 0 → t = a²; x = a → t = 0.

∫₀^a x√(a²−x²) dx = ½ ∫₀^(a²) t^(1/2) dt = ½ · (2/3) a³ = a³/3

Area = (4/a)(a³/3)

Area of the two loops = 4a²/3 square units (each loop has area 2a²/3).