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Triple integral x² over tetrahedron

Short answer

Tetrahedron bounded by x=0,y=0,z=0, x+y+z=a. Integrate in order z, y, x: I = ∫₀^a ∫₀^{a−x} ∫₀^{a−x−y} x² dz dy dx, which evaluates to a⁵/60.

Evaluate ∭ x² dx dy dz over the region v bounded by the planes x = 0, y = 0, z = 0 and x+y+z = a.

Region. The tetrahedron bounded by the coordinate planes and x + y + z = a: x ≥ 0, y ≥ 0, 0 ≤ z ≤ a − x − y, 0 ≤ y ≤ a − x, 0 ≤ x ≤ a.

I = ∫₀^a ∫₀^{a−x} ∫₀^{a−x−y} x² dz dy dx

Integrate in z

∫₀^{a−x−y} x² dz = x²(a − x − y).

Integrate in y

∫₀^{a−x} x²(a − x − y) dy = x² [ (a−x)y − y²/2 ]₀^{a−x}
                          = x² (a − x)²/2

Integrate in x

I = (1/2) ∫₀^a x²(a − x)² dx
  = (1/2) ∫₀^a (a²x² − 2a x³ + x⁴) dx
  = (1/2) ( a²·a³/3 − 2a·a⁴/4 + a⁵/5 )
  = (a⁵/2)(1/3 − 1/2 + 1/5)
  = (a⁵/2)(1/30)
  = a⁵/60

Answer: a⁵/60.

✓ Answer checked by Claude Opus 5.5 (high effort)

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