Skip to main content

Triple integral x dv over tetrahedron

Short answer

The region is the tetrahedron bounded by the coordinate planes and x+y+z=1. Integrate x dz dy dx with limits 0≤z≤1−x−y, 0≤y≤1−x, 0≤x≤1. Result: 1/24.

Evaluate: ∭_V x dv where V is bounded by the coordinate planes and the plane x+y+z = 1.

Region. The tetrahedron bounded by x = 0, y = 0, z = 0 and x + y + z = 1: 0 ≤ z ≤ 1 − x − y, 0 ≤ y ≤ 1 − x, 0 ≤ x ≤ 1.

I = ∫₀¹ ∫₀^{1−x} ∫₀^{1−x−y} x dz dy dx

z-integration

∫₀^{1−x−y} x dz = x(1 − x − y).

y-integration

∫₀^{1−x} x(1 − x − y) dy = x [ (1−x)y − y²/2 ]₀^{1−x}
                         = x (1−x)²/2

x-integration

I = (1/2) ∫₀¹ x(1 − x)² dx
  = (1/2) ∫₀¹ (x − 2x² + x³) dx
  = (1/2)(1/2 − 2/3 + 1/4)
  = (1/2)(1/12)
  = 1/24

(Check by Beta function: ∫₀¹ x(1−x)² dx = B(2,3) = 1!·2!/4! = 1/12.)

Answer: 1/24.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics chapter 2 questions →