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Evaluate using Gauss' divergence theorem F=xi+yj+z²k

Short answer

Divergence is 2 + 2z. Using cylindrical coordinates, ∭ (2+2z) r dz dr dθ over 0≤r≤3, 0≤θ≤2π, 0≤z≤2 equals 72π.

Using Gauss' divergence theorem, evaluate ∬_S F·n ds, where F = x i + y j + z² k and S is the surface bounded by the cylinder x² + y² = 9, z = 0 and z = 2.

Gauss' divergence theorem: ∬_S F·n dS = ∭_V div F dV, for the closed surface S enclosing V. Here S is the closed cylinder x² + y² = 9 with bottom z = 0 and top z = 2.

F = x i + y j + z² k
div F = ∂x/∂x + ∂y/∂y + ∂(z²)/∂z = 1 + 1 + 2z = 2 + 2z

Use cylindrical coordinates (x = r cos θ, y = r sin θ, dV = r dr dθ dz), with 0 ≤ r ≤ 3, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 2.

∭ (2 + 2z) dV = ∫₀^2π ∫₀³ ∫₀² (2 + 2z) r dz dr dθ
inner: ∫₀² (2 + 2z) dz = [2z + z²]₀² = 4 + 4 = 8
then:  ∫₀³ 8r dr = 8 · 9/2 = 36
then:  ∫₀^2π 36 dθ = 72π

Check by pieces: curved surface gives ∬(x² + y²)/3 dS = 3 · (2π·3·2) = 36π; top (z = 2, F·k = 4) gives 4·9π = 36π; bottom (z = 0) gives 0. Total 72π.

Answer: ∬_S F·n dS = 72π.

✓ Answer checked by Claude Opus 5.5 (high effort)

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