Since we have and Finally, the boundary term is obtained as follows: ∇ $_{y}$ϕ · ∇ y ( |∇ $_{y}$ϕ | 2 − |∇ $_{x}$ϕ | 2 ) = 2 γ $^{4}$ϕ$^{3}$d$_{2}$ ( ψ ) ∇ $_{y}$ψ · ∇ $_{y}$ψ + γ $^{3}$ϕ$^{3}$∇ $_{y}$ψ · ∇ y ( d$_{2}$ ( ψ )) ∇ $_{x}$ϕ · ∇ x ( |∇ $_{y}$ϕ | 2 − |∇ $_{x}$ϕ | 2 ) = 2 γ $^{4}$ϕ$^{3}$d$_{2}$ ( ψ ) ∇ $_{x}$ψ · ∇ $_{x}$ψ + γ $^{3}$ϕ$^{3}$∇ $_{x}$ψ · ∇ $_{x}$( d$_{2}$ ( ψ )) , J$_{6}$ = 2 s 3 ∫ Ω | z | 2 ∇ $_{y}$ϕ · ∇ y ( |∇ $_{y}$ϕ | 2 − |∇ $_{x}$ϕ | 2 ) dxdydt − 2 s 3 ∫ Ω | z | 2 ∇ $_{x}$ϕ · ∇ x ( |∇ $_{y}$ϕ | 2 − |∇ $_{x}$ϕ | 2 ) dxdydt = 4 ∫ Ω s $^{3}$γ$^{4}$ϕ3 | z | $^{2}$( d$_{2}$ ( ψ )) 2 dxdydt + Ω 2 s $^{3}$γ$^{3}$ϕ3 | z | 2 ∇ $_{y}$ψ · ∇ y ( d$_{2}$ ( ψ )) dxdydt − ∫ Ω 2 s $^{3}$γ$^{3}$ϕ3 | z | 2 ∇ $_{x}$ψ · ∇ $_{x}$( d$_{2}$ ( ψ )) dxdydt. (3.27) B$_{0}$ = − 2 Im ∫ Γ$_{y}$ sγϕz∂$_{t}$z ( ∇ $_{y}$ψ · ν ) dS$_{y}$dxdt +4 Re Γ$_{y}$ sγϕ ( ∂$_{ν}$z ) ( ∇ $_{x}$ψ · ∇ $_{x}$z − ∇ $_{y}$ψ · ∇ $_{y}$z ) dS$_{y}$dxdt + ∫ Γ$_{y}$ 2 sγϕ ( ∂$_{ν}$ψ ) ( |∇ $_{y}$z | 2 − |∇ $_{x}$z | 2 ) dS$_{y}$dxdt − Γ$_{y}$ 2 s $^{3}$γ$^{3}$ϕ$^{3}$∂$_{v}$ ψ | z | 2 d$_{2}$ ( ψ ) dS$_{y}$dxdt − ∫ Γ$_{y}$ 2 s ( γϕd$_{1}$ ( ψ ) + γ $^{2}$ϕd$_{2}$ ( ψ ) z ( ∂$_{ν}$z ) dS$_{y}$dxdt + Γ$_{y}$ s (( γ $^{2}$ϕd$_{1}$ ( ψ ) + γ $^{3}$ϕd$_{2}$ ( ψ ) ( ∂$_{ν}$ψ ) + γ $^{2}$ϕ ( ∂$_{ν}$ ( d$_{2}$ ( ψ ))) ) | z | 2 dS$_{y}$dxdt +2 Im ∫ Γ$_{x}$ sγϕz∂$_{t}$z ( ∇ $_{x}$ψ · ν ) dS$_{x}$dydt +4 Re Γ$_{x}$ sγϕ ( ∂$_{v}$z ) ( ∇ $_{y}$ψ · ∇ $_{y}$z − ∇ $_{x}$ψ · ∇ $_{x}$z ) dS$_{x}$dydt 15