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Volume of sphere by triple integration

Short answer

By symmetry, V = 8 times the first-octant volume. Integrate z from 0 to √(a²−x²−y²), y from 0 to √(a²−x²), x from 0 to a, giving V = (4/3)πa³.

Find by triple integration the volume of the sphere x²+y²+z² = a².

Method. By symmetry, V = 8 × (volume in the first octant), where 0 ≤ z ≤ √(a² − x² − y²), 0 ≤ y ≤ √(a² − x²), 0 ≤ x ≤ a.

V = 8 ∫₀^a ∫₀^{√(a²−x²)} ∫₀^{√(a²−x²−y²)} dz dy dx

z-integration

∫₀^{√(a²−x²−y²)} dz = √(a² − x² − y²).

y-integration

Put k² = a² − x² (constant for this step). Then

∫₀^k √(k² − y²) dy = [ (y/2)√(k²−y²) + (k²/2) sin⁻¹(y/k) ]₀^k
                   = (k²/2)(π/2) = πk²/4 = π(a² − x²)/4

x-integration

V = 8 ∫₀^a (π/4)(a² − x²) dx
  = 2π [ a²x − x³/3 ]₀^a
  = 2π (a³ − a³/3)
  = 4π a³/3

Answer: V = (4/3)π a³.

✓ Answer checked by Claude Opus 5.5 (high effort)

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