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Prove triple integral of √(x²+y²+z²)e^−(x²+y²+z²) is 2π

Short answer

Switch to spherical coordinates: integrand r·e^{-r²}·r² sinφ dr dφ dθ. Angle integrals give 4π, radial integral ∫₀∞ r³ e^{-r²} dr = 1/2, product is 2π.

Prove that ∫(−∞)^∞ ∫(−∞)^∞ ∫_(−∞)^∞ √(x²+y²+z²) e^−(x²+y²+z²) dx dy dz = 2π. [Integrand as read from scan; the √ factor is unclear]

Reading. The integrand is read as √(x² + y² + z²) · e^−(x² + y² + z²), the √ factor being the radius. This matches the stated value 2π.

I = ∭ over all space √(x² + y² + z²) e^−(x²+y²+z²) dx dy dz.

Spherical polar coordinates

x = r sin φ cos θ, y = r sin φ sin θ, z = r cos φ, dx dy dz = r² sin φ dr dφ dθ. Since x² + y² + z² = r², the whole space is covered by 0 ≤ r < ∞, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π.

I = ∫₀^{2π} dθ ∫₀^π sinφ dφ ∫₀^∞ r · e^(−r²) · r² dr
  = (2π)(2) ∫₀^∞ r³ e^(−r²) dr

The radial integral

Put t = r², dt = 2r dr, so r³ dr = (1/2) t dt:

∫₀^∞ r³ e^(−r²) dr = (1/2) ∫₀^∞ t e^(−t) dt = (1/2) Γ(2) = 1/2

Result

I = 4π · (1/2) = 2π

Hence the integral equals 2π. ∎

✓ Answer checked by Claude Opus 5.5 (high effort)

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