I was reading this paper and I think that I find a mistake, may be I'm wrong, but I want to be sure.
They take the variation with respect to the metric $g_{\alpha\beta}$ of this function
$$S(\delta \Omega)=\int_{\delta \Omega}n_{\nu}s^{\nu}\sqrt{h}d^{d-1}x$$
With some fixed boundary conditions $g_{\alpha\beta}(\delta \Omega)=g_{\alpha\beta}^{\delta \Omega}$. $s^{\nu}$ is a function that depends of the metric and $\delta\Omega$ is a Jordan Orientable surface with normal $n_{\nu}$.
They define a family of metrics
$$g_{\alpha\beta}(x^{\mu})=g*_{\alpha\beta}(\mu)+\delta_{\epsilon}(g_{\alpha\beta})x^{\mu} $$
Where $g*_{\alpha\beta}$ is the metric that extremize $S(\delta\Omega)$, $\epsilon\in R$. $\delta_{\epsilon}(g_{\alpha\beta})$ satisfices the boundary condition $\delta_{\epsilon}(g_{\alpha\beta})(\delta\Omega)=0$ and $\lim_{\epsilon \rightarrow 0} \delta_{\epsilon}(g_{\alpha\beta})(x^{\mu})=0$.
The the variation with respect to the metric of the first equation is
$$\lim_{\epsilon \rightarrow 0} \frac{\delta_{\epsilon(S)(\delta \Omega)}}{\epsilon}=0$$
$$\lim_{\epsilon \rightarrow 0} \frac{\delta_{\epsilon(S)(\delta \Omega)}}{\epsilon}=\int_{\delta \Omega}n_{\nu}\lim_{\epsilon \rightarrow 0}\frac{\delta_{\epsilon}(s^{\nu})}{\epsilon}\sqrt{h}d^{d-1}x=0$$ But I think that the last equation is wrong. I think that we also have to take the variation with respect the metric of the normal $n_{\nu}$, something like this: $$\lim_{\epsilon \rightarrow 0} \frac{\delta_{\epsilon(S)(\delta \Omega)}}{\epsilon}=\int_{\delta \Omega}\lim_{\epsilon \rightarrow 0}\frac{\delta_{\epsilon}(n_{\nu}s^{\nu})}{\epsilon}\sqrt{h}d^{d-1}x=0$$
$$n_{\alpha}=\frac{\partial_{\alpha}f}{\sqrt{|g^{\alpha \beta}\partial_{\alpha}f \partial_{\beta}f | }}$$
$$\textbf{EDIT}$$ I'm not looking for the complete way of taking the variation of this function. I'm looking for and answer that say's if I have to do something like this $\delta_{\epsilon}n_{\nu}(s^{\nu})$ or like this $\delta_{\epsilon}(n_{\nu}s^{\nu})$ in the variation.