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Area between y²(2a−x)=x³ and its asymptote

Short answer

Asymptote is x=2a. Area = 2∫₀²ᵃ x^(3/2)/√(2a−x) dx. Substituting x=2a sin²θ reduces to 16a²∫₀^{π/2} sin⁴θ dθ = 3πa².

Find the area of the curve y²(2a−x) = x³ and its asymptotes.

Curve: y²(2a − x) = x³ (cissoid), so y² = x³/(2a − x). It is symmetrical about the x-axis, exists for 0 ≤ x < 2a, passes through the origin and has the asymptote x = 2a (y → ±∞ as x → 2a).

The area between the curve and the asymptote is twice the area above the x-axis:

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

Put x = 2a sin²θ, dx = 4a sin θ cos θ dθ, 2a − x = 2a cos²θ. Limits 0 to π/2.

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

(using ∫₀^(π/2) sin⁴θ dθ = (3/4)(1/2)(π/2) = 3π/16)

Area = 3πa².

✓ Answer checked by Claude Opus 5.5 (high effort)

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