Evaluate using Gauss' divergence theorem F=xi−yj+(z²−1)k

Gauss' divergence theorem: ∬_S F·n dS = ∭_V ∇·F dV.

F = x i − y j + (z² − 1) k. S is the closed cylinder x² + y² = 4 with the base z = 0 and the top z = 1.

Divergence

∇·F = ∂x/∂x + ∂(−y)/∂y + ∂(z² − 1)/∂z = 1 − 1 + 2z = 2z

Volume integral

Use cylindrical coordinates: x = r cos θ, y = r sin θ, dV = r dr dθ dz; r from 0 to 2, θ from 0 to 2π, z from 0 to 1.

∭ 2z dV = ∫₀¹ 2z dz · ∫₀^{2π} dθ · ∫₀² r dr
        = [z²]₀¹ · 2π · [r²/2]₀²
        = 1 · 2π · 2 = 4π

Answer: ∬_S F·n dS = 4π.

Check by surfaces

• Top (z = 1, n = k): F·n = z² − 1 = 0 ⇒ contributes 0.
• Base (z = 0, n = −k): F·n = −(z² − 1) = 1 ⇒ contributes the area 4π.
• Curved surface: n = (x i + y j)/2, F·n = (x² − y²)/2, whose integral over the cylinder is 0 by symmetry.

Total = 4π ✓.