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Verify divergence theorem, quarter hemisphere

Short answer

The volume integral of ∇·A over the region is 81π/2+9, matching the total flux through the bounding surfaces, verifying the theorem.

Verify the divergence theorem (evaluate both sides of the divergence theorem) for the function A = r² a_r + r sinθ cosφ a_θ over the surface of a quarter of a hemisphere defined by 0 < r < 3, 0 < φ < π/2, 0 < θ < π/2.

Divergence theorem: ∮ A · dS = ∫ (∇ · A) dv.

Given: A = r² a_r + r sinθ cosφ a_θ. Region: 0 < r < 3, 0 < θ < π/2, 0 < φ < π/2 (one-eighth of a sphere).

Volume integral

∇ · A = (1/r²) ∂(r² A_r)/∂r + (1/(r sinθ)) ∂(sinθ A_θ)/∂θ + (1/(r sinθ)) ∂A_φ/∂φ

(1/r²) ∂(r⁴)/∂r = 4r
(1/(r sinθ)) ∂(r sin²θ cosφ)/∂θ = (1/(r sinθ)) (2r sinθ cosθ cosφ) = 2 cosθ cosφ

∇ · A = 4r + 2 cosθ cosφ

∫ (∇ · A) dv = ∫₀^(π/2) ∫₀^(π/2) ∫₀³ (4r + 2cosθ cosφ) r² sinθ dr dθ dφ

Term 1: 4 ∫₀³ r³ dr · ∫₀^(π/2) sinθ dθ · ∫₀^(π/2) dφ = 4 (81/4)(1)(π/2) = 81π/2 = 127.23
Term 2: 2 ∫₀³ r² dr · ∫₀^(π/2) sinθ cosθ dθ · ∫₀^(π/2) cosφ dφ = 2 (9)(1/2)(1) = 9

Total = 81π/2 + 9 = 136.23

Surface integral over the four faces

1. Curved face r = 3 (dS = r² sinθ dθ dφ a_r):
   Ψ₁ = ∫∫ (3²)(3²) sinθ dθ dφ = 81 (1)(π/2) = 81π/2 = 127.23

2. Face θ = π/2 (outward normal +a_θ, dS = r sinθ dr dφ a_θ):
   A_θ = r cosφ at θ = π/2
   Ψ₂ = ∫₀^(π/2)∫₀³ (r cosφ)(r) dr dφ = (9)(1) = 9

3. Faces φ = 0 and φ = π/2 (normal ±a_φ): A has no a_φ component → 0

4. The face r = 0 is a point, θ = 0 is the z-axis line → 0

Ψ = 127.23 + 9 = 136.23

Answer: Both sides equal 81π/2 + 9 ≈ 136.2. The divergence theorem is verified.

✓ Answer checked by Claude Opus 5.5 (high effort)

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