Skip to main content

Volume of ellipsoid by triple integration

Short answer

Using symmetry, V = 8 × first-octant volume. Integrate in order z, y, x with limits dependent on the ellipsoid equation to obtain V = (4/3)πabc.

Find by triple integration the volume of the ellipsoid x²/a² + y²/b² + z²/c² = 1.

Method. The ellipsoid x²/a² + y²/b² + z²/c² = 1 is symmetric about all three coordinate planes, so V = 8 × (first octant volume).

In the first octant: 0 ≤ x ≤ a, 0 ≤ y ≤ b√(1 − x²/a²), 0 ≤ z ≤ c√(1 − x²/a² − y²/b²).

V = 8 ∫₀^a ∫₀^{b√(1−x²/a²)} ∫₀^{c√(1−x²/a²−y²/b²)} dz dy dx

z-integration

The z-integral gives c√(1 − x²/a² − y²/b²) = (c/b)√(k² − y²), where k² = b²(1 − x²/a²).

y-integration

(c/b) ∫₀^k √(k² − y²) dy = (c/b)(πk²/4)
                         = (c/b)(π/4) b²(1 − x²/a²)
                         = (π b c/4)(1 − x²/a²)

x-integration

V = 8 (π b c/4) ∫₀^a (1 − x²/a²) dx
  = 2π b c [ x − x³/(3a²) ]₀^a
  = 2π b c (2a/3)
  = 4π abc/3

Answer: V = (4/3)π abc. For a = b = c this reduces to the sphere (4/3)π a³.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 2 questions →