Verify Gauss' divergence theorem F=4xi−2y²j+z²k

Gauss' theorem: ∬_S F·n dS = ∭_V div F dV. F = 4x i − 2y² j + z² k over the cylinder x² + y² ≤ 4, 0 ≤ z ≤ 3.

Volume integral

div F = 4 − 4y + 2z
∭ (4 − 4y + 2z) dV with cylindrical coordinates (r ≤ 2, 0 ≤ z ≤ 3):
 4 ∭ dV = 4 · π(4)(3) = 48π
 −4 ∭ y dV = 0      (y = r sin θ, integral over θ is 0)
 2 ∭ z dV = 2 · (π·4) · ∫₀³ z dz = 2 · 4π · 9/2 = 36π
Total = 84π

Surface integral

• Top z = 3, n = k: F·n = z² = 9. Flux = 9 · 4π = 36π.
• Bottom z = 0, n = −k: F·n = −z² = 0. Flux = 0.
• Curved surface: n = (x i + y j)/2, F·n = (4x² − 2y³)/2. With x = 2cos θ, y = 2sin θ, dS = 2 dθ dz:

∫₀³ ∫₀^2π (4x² − 2y³) dθ dz = 3 ∫₀^2π (16cos²θ − 16 sin³θ) dθ
                             = 3 (16π − 0) = 48π

Total = 36π + 0 + 48π = 84π, equal to the volume integral.

So Gauss' theorem is verified.