Average J over surface and ∂ρv/∂t

Given J = 20 sinθ cosφ a_r + (1/r) a_θ A/m² in spherical coordinates.

(i) Average value of J_r over r = 1, 0 < θ < π/2, 0 < φ < π/2

Average = (∫ J_r dS) / (∫ dS), with dS = r² sinθ dθ dφ (r = 1).

∫ J_r dS = ∫₀^(π/2) ∫₀^(π/2) 20 sinθ cosφ · sinθ dθ dφ
         = 20 ∫₀^(π/2) sin²θ dθ · ∫₀^(π/2) cosφ dφ
         = 20 · (π/4) · 1
         = 5π = 15.71

∫ dS = ∫₀^(π/2) sinθ dθ ∫₀^(π/2) dφ = (1)(π/2) = π/2 = 1.571

J_r(avg) = 5π/(π/2) = 10 A/m²

(ii) ∂ρ_v/∂t

From the continuity equation, ∇·J = −∂ρ_v/∂t.

∇·J = (1/r²) ∂(r² J_r)/∂r + (1/(r sinθ)) ∂(sinθ J_θ)/∂θ + (1/(r sinθ)) ∂J_φ/∂φ

term 1: (1/r²) ∂(20 r² sinθ cosφ)/∂r = 40 sinθ cosφ / r
term 2: (1/(r sinθ)) ∂(sinθ · 1/r)/∂θ = cosθ/(r² sinθ)
term 3: 0

∇·J = 40 sinθ cosφ / r + cosθ/(r² sinθ)

∂ρ_v/∂t = −∇·J = −[ 40 sinθ cosφ / r + cosθ/(r² sinθ) ]  A/m³ (C/m³ per second)

Answers: (i) average J_r = 10 A/m²; (ii) ∂ρ_v/∂t = −(40 sinθ cosφ)/r − cosθ/(r² sinθ) A/m³.