In (3.15), we used the equality Re ( iz ) = Im (¯ z ) . In (3.16), we used the equality Re z$_{y}$$_{j}$ ¯ z$_{y}$$_{i}$$_{y}$$_{j}$ = 1 2 ( ∣ z$_{y}$$_{j}$ ∣ 2 ) y$_{i}$ . I$_{3}$ = − 4 s Re ∫ Ω Δ$_{y}$ z ∇ $_{y}$ϕ · ∇ $_{y}$zdxdydt + 4 s Re ∫ Ω Δ$_{y}$ z ∇ $_{x}$ϕ · ∇ $_{x}$zdxdydt = 4 s Re ∫ Ω n ∑ i,j =1 ϕ$_{y}$$_{i}$$_{y}$$_{j}$z$_{y}$$_{i}$z$_{y}$$_{j}$dxdydt − 2 s ∫ Ω Δ$_{y}$ ϕ |∇ $_{y}$z | 2 dxdydt +2 s Γ$_{y}$ ( ∂$_{ν}$ϕ ) |∇ $_{y}$z | 2 dS$_{y}$dxdt − 4 s Re Γ$_{y}$ ( ∂$_{ν}$z ) ∇ $_{y}$ϕ · ∇ $_{y}$zdS$_{y}$ dxdt − 4 s Re ∫ Ω m ∑ j =1 n ∑ i =1 ϕ$_{x}$$_{j}$$_{y}$$_{i}$z$_{x}$$_{j}$ z$_{y}$$_{i}$dxdydt + 2 s ∫ Ω Δ$_{x}$ ϕ |∇ $_{y}$z | 2 dxdydt − 2 s ∫ Γ$_{x}$ ( ∂$_{ν}$ϕ ) |∇ $_{y}$z | 2 dS$_{x}$dydt +4 s Re ∫ Γ$_{y}$ ( ∂$_{ν}$z ) ∇ $_{x}$ϕ · ∇ $_{x}$zdS$_{y}$ dxdt. (3.16) I$_{4}$ = − 2 s Re ∫ Ω Δ$_{y}$ z (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) zdxdydt = − s Re Ω Δ$_{y}$ (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) | z | 2 dxdydt +2 s ∫ Ω (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) |∇ $_{y}$z | 2 dxdydt + s Γ$_{y}$ ∂$_{ν}$ (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) | z | 2 dS$_{y}$dxdt − 2 s Re ∫ Γ$_{y}$ ( ∂$_{ν}$z ) (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) zdS$_{y}$dxdt. (3.17) 9