Verify Gauss' divergence theorem F=yi+xj+z²k

Gauss' theorem: ∬_S F·n dS = ∭_V div F dV, with F = y i + x j + z² k and V the cylinder x² + y² ≤ a², 0 ≤ z ≤ h.

Volume integral

div F = 0 + 0 + 2z = 2z
∭ 2z dV = ∫₀^h 2z (πa²) dz = πa² h²

(the cross-section at height z has area πa²).

Surface integral

• Top z = h, n = k: F·n = z² = h². Flux = h² · πa² = πa²h².
• Bottom z = 0, n = −k: F·n = −z² = 0. Flux = 0.
• Curved surface: n = (x i + y j)/a, so F·n = (xy + yx)/a = 2xy/a. With x = a cos θ, y = a sin θ, dS = a dθ dz:

∫₀^h ∫₀^2π (2a² sin θ cos θ / a) a dθ dz = a² h ∫₀^2π sin 2θ dθ = 0

Total surface flux = πa²h² + 0 + 0 = πa²h², equal to the volume integral.

So Gauss' divergence theorem is verified.