Area between y²(a−x)=x³ and its asymptote

Short answer

The asymptote is x=a. Area = 2∫₀^a x^(3/2)(a−x)^(−1/2) dx. Substitute x = a sin²θ to obtain 4a² ∫₀^{π/2} sin⁴θ dθ = 3πa²/4.

Find the area enclosed by y²(a−x) = x³ and its asymptotes.

Curve: y²(a − x) = x³ (the cissoid of Diocles). Then y² = x³/(a − x).

The curve exists for 0 ≤ x < a (for x < 0 or x > a the right side is negative).
It is symmetric about the x-axis, passes through the origin (a cusp), and as x → a, y → ±∞. So x = a is the asymptote.

The required area lies between the curve and the asymptote x = a, so it is twice the area above the x-axis:

A = 2 ∫₀^a y dx = 2 ∫₀^a x^(3/2) (a − x)^(−1/2) dx

Put x = a sin²θ, dx = 2a sin θ cos θ dθ, a − x = a cos²θ. Limits: 0 to π/2.

A = 2 ∫₀^(π/2) a^(3/2) sin³θ · a^(−1/2) (1/cosθ) · 2a sinθ cosθ dθ
  = 4a² ∫₀^(π/2) sin⁴θ dθ
  = 4a² · (3/4)(1/2)(π/2)
  = 4a² · 3π/16

Area = 3πa²/4.

✓ Answer checked by Claude Opus 5.5 (high effort)

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