Change order: y²/√(y⁴−a²x²)

How I read the limits. As printed, the integral is I = ∫₀^a ∫_{√(ax)}^{a} y²/√(y⁴ − a²x²) dy dx. This reading is consistent: the integrand is real exactly when y⁴ ≥ a²x², i.e. y ≥ √(ax), which is the lower limit of y.

Region

0 ≤ x ≤ a and √(ax) ≤ y ≤ a. The lower boundary y = √(ax) is the parabola y² = ax, so x ≤ y²/a. Taking horizontal strips: y from 0 to a, and x from 0 to y²/a.

y
a |- - - - - -*(a,a)
  |         . '
  |     . '   region: between x=0
  |  . '      and parabola y² = ax
  |_'_ _ _ _ _ _ x
 0               a

Reversed order

I = ∫₀^a y² [ ∫₀^{y²/a} dx / √(y⁴ − a²x²) ] dy

Inner: ∫ dx/√(y⁴ − a²x²) = (1/a) sin⁻¹(ax/y²). Between 0 and y²/a: (1/a)(sin⁻¹ 1 − 0) = π/(2a).

I = ∫₀^a y² · π/(2a) dy
  = (π/2a)(a³/3)
  = π a²/6

Answer: π a²/6.