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Surface integral over a surface F=4xi−2y²j+z²k

Short answer

Using Gauss divergence theorem, ∇·F = 4 − 4y + 2z. Volume integral in cylindrical coordinates over cylinder radius 2, height 3 gives 84π.

Evaluate ∬_S F·n ds, where F = 4x i − 2y² j + z² k taken over the region bounded by the cylinder x² + y² = 4 and the planes z = 0, z = 3.

F = 4x i − 2y² j + z² k, taken over the closed surface bounded by the cylinder x² + y² = 4 and the planes z = 0, z = 3. Use Gauss' divergence theorem: ∬_S F·n dS = ∭_V ∇·F dV.

Divergence

∇·F = ∂(4x)/∂x + ∂(−2y²)/∂y + ∂(z²)/∂z = 4 − 4y + 2z

Volume integral (cylindrical coordinates)

x = r cos θ, y = r sin θ, dV = r dr dθ dz; 0 ≤ r ≤ 2, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ 3.

∭ (4 − 4y + 2z) dV = ∭ 4 dV − 4∭ y dV + 2∭ z dV
  • ∭ 4 dV = 4 × (volume) = 4 × π(2)²(3) = 48π
  • ∭ y dV = 0, because it contains ∫₀^{2π} sin θ dθ = 0
  • 2∭ z dV = 2 · (∫₀³ z dz)(∫₀^{2π} dθ ∫₀² r dr) = 2 · (9/2)(4π) = 36π
Total = 48π − 0 + 36π = 84π

Answer: ∬_S F·n dS = 84π.

✓ Answer checked by Claude Opus 5.5 (high effort)

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