This amounts to proving that

(6.24)

$$\int_{\mathcal{M}_{T}}\mathscr{F}(\sqrt{-1}\partial\bar{\partial}\phi)\omega_{T}^{n}=0.$$

By Stokes’ theorem,

(6.25)

$$\int_{\mathcal{M}_{T}}(\omega_{T}+\sqrt{-1}\partial\bar{\partial}\phi)^{n}-\int_{\mathcal{M}_{T}}\omega_{T}^{n}=\int_{\partial\mathcal{M}_{T}}\gamma,$$

where $\gamma$ is the sum of terms involving one factor $d^{c}\phi$ and either $dd^{c}\phi$ or $\omega_{T}$. We claim that $\gamma$ identically vanishes on $\partial\mathcal{M}_{T}$. It suffices to show $\partial_{t}\lrcorner\gamma=0$. Since by assumption $\phi$ is $S^{1}$-invariant, so we have $\partial_{t}\phi=0$. By the Neumann boundary condition, we also have $d^{c}\phi(\partial_{t})=0$ on $\partial\mathcal{M}_{T}$. This follows from the observation that $J\partial_{t}=\nabla z$. Now

(6.26)		$\displaystyle\partial_{t}\lrcorner\omega_{T}|_{\partial\mathcal{M}_{T}}$	$\displaystyle=dz|_{\partial\mathcal{M}_{T}}=0,$
(6.27)		$\displaystyle\partial_{t}\lrcorner dd^{c}\phi$	$\displaystyle=\mathcal{L}_{\partial_{t}}(d^{c}\phi)-d(\partial_{t}\lrcorner d^{c}\phi)=-d(\partial_{t}\lrcorner d^{c}\phi).$

The last term vanishes on $\partial\mathcal{M}_{T}$ since $d^{c}\phi(\partial_{t})=0$ pointwise on $\partial\mathcal{M}_{T}$. ∎