Verify divergence theorem, D field

Divergence theorem: The net outward flux of D through a closed surface equals the volume integral of ∇ · D over the enclosed volume: ∮ D · dS = ∫ (∇ · D) dv.

Given: D = (2cosθ/r³) a_r + (sinθ/r³) a_θ C/m². Region: 1 < r < 2, 0 < θ < π/2, 0 < φ < π/2.

Right side: volume integral

∇ · D = (1/r²) ∂(r² D_r)/∂r + (1/(r sinθ)) ∂(sinθ D_θ)/∂θ + (1/(r sinθ)) ∂D_φ/∂φ

(D has no φ-component, so the last term is zero.)

(1/r²) ∂(r² · 2cosθ/r³)/∂r = (1/r²) ∂(2cosθ/r)/∂r = −2cosθ/r⁴
(1/(r sinθ)) ∂(sinθ · sinθ/r³)/∂θ = (1/(r sinθ)) (2 sinθ cosθ / r³) = 2cosθ/r⁴

∇ · D = −2cosθ/r⁴ + 2cosθ/r⁴ = 0

So ∫ (∇ · D) dv = 0.

Left side: closed surface integral

The region has five faces.

1. Outer, r = 2:   dS = r² sinθ dθ dφ a_r,  D_r = 2cosθ/8
   Ψ = ∫₀^(π/2)∫₀^(π/2) (2cosθ/8)(4) sinθ dθ dφ = ∫ cosθ sinθ dθ · ∫dφ = (1/2)(π/2) = π/4

2. Inner, r = 1:   dS = −r² sinθ dθ dφ a_r,  D_r = 2cosθ
   Ψ = −∫∫ 2cosθ sinθ dθ dφ = −(1)(π/2) = −π/2

3. Face θ = π/2 (part of the xy-plane): outward normal is +a_θ, dS = r sinθ dr dφ a_θ
   D_θ = 1/r³ at θ = π/2:
   Ψ = ∫₀^(π/2)∫₁² (1/r³)(r) dr dφ = (1/2)(π/2) = π/4

4. θ = 0: this is only the z-axis (zero area): Ψ = 0

5. φ = 0 and φ = π/2 faces: D has no a_φ component, so D · dS = 0.

Total flux:

Ψ = π/4 − π/2 + π/4 + 0 = 0

Answer: Both sides equal 0, so the divergence theorem is verified. (D is solenoidal here: the same flux enters through r = 1 as leaves through r = 2 and θ = π/2.)