Volume about the x-axis for the astroid is

2π integral y^2 dx from 0 to a

The integral reduces to

6π a^3 integral sin^7 t cos^2 t dt

6π a^3 integral sin^6 t cos^2 t dt

3π a^3 integral sin^7 t cos^2 t dt

6π a^3 integral sin^2 t cos^7 t dt

Short answer

Parametrize with x=a cos³t, y=a sin³t. The volume becomes 6πa³∫₀^{π/2} sin⁷t cos²t dt, giving 32πa³/105 cubic units.

Find the volume generated by revolution of astroid x^(2/3) + y^(2/3) = a^(2/3) about x-axis.

Astroid: x^(2/3) + y^(2/3) = a^(2/3). The solid is symmetrical about the y-axis, so V = 2π ∫₀^a y² dx.

Use x = a cos³t, y = a sin³t, dx = −3a cos²t sin t dt. As x goes 0 → a, t goes π/2 → 0.

V = 2π ∫_(π/2)^0 a² sin⁶t (−3a cos²t sin t) dt
  = 6π a³ ∫₀^(π/2) sin⁷t cos²t dt

Beta function: ∫₀^(π/2) sin⁷t cos²t dt = Γ(4) Γ(3/2) / (2Γ(11/2)).

Γ(4) = 6; Γ(3/2) = √π/2; Γ(11/2) = (9/2)(7/2)(5/2)(3/2)(1/2)√π = 945√π/32.

= 6 · (√π/2) / (2 · 945√π/32) = 3 · 32/(2 · 945) = 48/945 = 16/315.

V = 6πa³ · 16/315 = 96πa³/315

V = 32πa³/105 cubic units.

✓ Answer checked by Claude Opus 5.5 (high effort)