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Evaluate using Gauss' divergence theorem F=x²i+zj+yzk

Short answer

Divergence of x² i + z j + yz k is 2x + y. Volume integral over the unit cube gives 3/2.

Apply Gauss divergence theorem to evaluate ∬_S F·n ds where F = x² i + z j + yz k taken over the cube bounded by x = 0, x = 1, y = 0, y = 1, z = 0, z = 1.

F = x² i + z j + yz k, over the closed surface of the unit cube 0 ≤ x, y, z ≤ 1.

Divergence

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

Volume integral

∬_S F·n dS = ∭_V (2x + y) dV
           = ∫₀¹∫₀¹∫₀¹ (2x + y) dx dy dz
           = ∫₀¹∫₀¹ [x² + xy]₀¹ dy dz
           = ∫₀¹∫₀¹ (1 + y) dy dz
           = ∫₀¹ [y + y²/2]₀¹ dz
           = 3/2

Answer: ∬_S F·n dS = 3/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

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