In (3.17), we used the equality Re z ∇ $_{y}$z = 1 2 ∇ y ( | z | 2 ) . In (3.18), we used the equality Re z$_{x}$$_{j}$ z$_{x}$$_{j}$$_{y}$$_{i}$ = 1 2 ( ∣ z$_{x}$$_{j}$ ∣ 2 ) y$_{i}$ . I$_{5}$ = 4 s Re ∫ Ω Δ$_{x}$ z ∇ $_{y}$ϕ · ∇ $_{y}$zdxdydt − 4 s Re ∫ Ω Δ$_{x}$ z ∇ $_{x}$ϕ · ∇ $_{x}$zdxdydt = − 4 s Re ∫ Ω m ∑ j =1 n ∑ i =1 ϕ$_{y}$$_{i}$$_{x}$$_{j}$ z$_{y}$$_{i}$z$_{x}$$_{j}$dxdydt − 2 s ∫ Ω m ∑ i =1 Δ$_{y}$ ϕ |∇ $_{x}$z | 2 dxdydt − 2 s ∫ Γ$_{y}$ ( ∂$_{ν}$ϕ ) |∇ $_{x}$z | 2 dS$_{y}$dxdt + 4 s Re ∫ Γ$_{x}$ ( ∂$_{ν}$z ) ∇ $_{y}$ϕ · ∇ $_{y}$zdS$_{x}$ dydt +4 s Re ∫ Ω n ∑ i,j =1 ϕ$_{x}$$_{i}$$_{y}$$_{j}$ z$_{x}$$_{i}$z$_{x}$$_{j}$dxdydt − 2 s Ω Δ$_{x}$ ϕ |∇ $_{x}$z | 2 dxdydt − 2 s ∫ Γ$_{x}$ ( ∂$_{ν}$ϕ ) |∇ $_{x}$z | 2 dS$_{x}$dydt +4 s Re ∫ Γ$_{x}$ ( ∂$_{ν}$z ) ∇ $_{x}$ϕ · ∇ $_{x}$zdS$_{x}$ dydt. (3.18) I$_{6}$ = 2 s Re ∫ Ω Δ$_{x}$ z (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) zdxdydt = − 2 s Ω (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) |∇ $_{x}$z | 2 dxdydt + s ∫ Ω Δ$_{x}$ (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ )) | z | 2 dxdydt − s Γ$_{x}$ ∂$_{ν}$ (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) | z | 2 dS$_{x}$dydt +2 s Re ∫ Γ$_{x}$ ( ∂$_{ν}$z ) (Δ$_{y}$ ϕ − Δ$_{x}$ ϕ ) zdS$_{x}$dydt. (3.19) 10