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Evaluate using Gauss' divergence theorem

Short answer

Gauss theorem equates flux to volume integral of divergence. For F = (x³−yz)i − 2x²y j + 2k, ∇·F = x². Integrating over the cube 0 ≤ x,y,z ≤ a gives a⁵/3.

State Gauss divergence theorem in vector calculus. Apply it to evaluate ∬_S [(x³ − yz) i − 2x²y j + 2k]·n ds where S denotes the surface of the cube bounded by the planes x = 0, x = a, y = 0, y = a, z = 0, z = a.

Gauss' divergence theorem

Let S be a closed surface enclosing a volume V, with n the outward unit normal. If F has continuous first partial derivatives in V, then

∬_S F·n dS = ∭_V (∇·F) dV

The outward flux through a closed surface equals the integral of the divergence over the volume enclosed.

Application

F = (x³ − yz) i − 2x²y j + 2k, and S is the closed surface of the cube 0 ≤ x, y, z ≤ a.

Divergence

∇·F = ∂(x³ − yz)/∂x + ∂(−2x²y)/∂y + ∂(2)/∂z = 3x² − 2x² + 0 = x²

Volume integral

∭_V x² dV = ∫₀ᵃ ∫₀ᵃ ∫₀ᵃ x² dx dy dz = (a³/3)(a)(a) = a⁵/3

Answer: ∬_S F·n dS = a⁵/3.

✓ Answer checked by Claude Opus 5.5 (high effort)

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