{R}}^{2}\to{\mathbb{R}}$ defined on an open subset $\Omega$, we define the graph $G_u:=\{(z,t): z\in\Omega, t=u(z)\}$. By \eqref{eq:area}, the sub-Riemannian area of the graph is given by
\[
A(G_u)=\int_{G_u} |N_h|\,dG_u,
\]
where $dG_u$ is the Riemannian metric of the graph and $|N_h|$ is the modulus of the horizontal projection of a unit normal to $G_u$. We consider on $\Omega$ the basis of vector~fields $
\big\{\tfrac{\partial}{\partial x},\tfrac{\partial}{\partial y}\big\}$.

By the Riemannian area formula
\begin{equation}
\label{eq:dgu}
dG_u=\text{Jac}\,d\mathcal{L}^{2},
\end{equation}
where $d\mathcal{L}^{2}$ is Lebesgue measure in ${\mathbb{R}}^{2}$ and $\text{Jac}$ is the Jacobian of the canonical map $\Omega\to G_u$ given by
\begin{equation}
\label{eq:jac}
\text{Jac}=\{\det(g)+g_{11}(u_y+x)^2+g_{22}(u_x-y)^2-2g_{12}(u_x-y)(u_y+x)\}^{1/2}
\end{equation}
where $g$ is the matrix of the metric, with elements $g_{ij}:=g(Z_i,Z_j)$. 

Let us compute the composition of $|N_h|$ with the map $\Omega\to G_u$. The tangent space $T G_u$ is spanned by
\begin{equation}
\label{eq:zi}
X+(u_x-y)\,T, \quad Y+(u_y+x)\,T.
\end{equation}
So the projection to $\Omega$ of the singular set $(G_u)_0$ is the set $\Omega_0\subset\Omega$ defined by $\Omega_0:=\{z\in\Omega: (\nabla u +F)=0\}$, where $\nabla$ is the Euclidean gradient in ${\mathbb{R}}^2$. Let us compute a \emph{downward pointing} normal vector $\tilde{N}$ to $G_u$ writing
\begin{equation}
\label{eq:tilden}
\tilde{N}=\sum_{i=1}^{2}(a_i Z_i)-T. 
\end{equation}
The horizontal component of $\tilde{N}$ is $\tilde{N}_h=\sum_{i=1}^{2} a_iZ_i$.  We have
\[
\sum_{i=1}^{2} a_ig_{ij}=g(\tilde{N}_h,Z_j)=g(\tilde{N},Z_j)=-(\nabla u+F)_j\escpr{\tilde{N},T}=(\nabla u +F)_j,
\]
since $Z_j$ is horizontal, $\tilde{N}$ is orthogonal to $Z_j$ defined by \eqref{eq:zi}, and \eqref{eq:tilden}. Hence 
\[
(a_1, a_{2})=b(\nabla u+F),
\]
where $b$ is the inverse of the matrix $\{g_{ij}\}_{i,j=1,2}$. So we get
\begin{equation}
\label{eq:modtilden}
|\tilde{N}|=(1+\escpr{\nabla u +F,b(\nabla u +F)})^{1/2},
\end{equation}
and
\[
|\tilde{N}_h|=\escpr{\nabla u+F, b(\nabla u+F)}^{1/2},
\]
where $\escpr{,}$ is the Euclidean Riemannian metric in ${\mathbb{R}}^{2}$, and so
\begin{equation}
\label{eq:modnh}
|N_h|=\frac{|\tilde{N}_h|}{|\tilde{N}|}=\frac{\escpr{\nabla u+F, b(\nabla u+F)}^{1/2}}{\big(1+\escpr{\nabla u +F,b(\nabla u +F)}\big)^{1/2}}.
\end{equation}
Observe that, from \eqref{eq:tilden} and \eqref{eq:modtilden} we also get
\begin{equation}
\label{eq:nxt}
g(N,T)=-\frac{1}{\big(1+\escpr{\nabla u+F,b(\nabla  u+F)}\big)^{1/2}}.
\end{equation}
Hence we obtain from \eqref{eq:area}, \eqref{eq:dgu}, \eqref{eq:jac} and \eqref{eq:modnh}
\begin{equation}
\label{eq:agu}
A(G_u)=\int_\Omega \escpr{\nabla u+F, b(\nabla u+F)}^{1/2}\,\frac{\det(g_{ij}+(\nabla u+F)_i(\nabla u+F)_j)^{1/2}}{\big(1+\escpr{\nabla u +F,b(\nabla u +F)}\big)^{1/2}}
\,d\mathcal{L}^{2}.
\end{equation}

We can compute the mean curvature of a graph $G_u$ \cite[Lemma~4.2]{Ga-Ri}

\begin{lemma}
\label{1variation:graph} 
Let us consider the contact sub-Riemannian manifold $(\mathbb{R}^{3},g_{\mathcal{H}_0},\omega_0)$, where $\omega_0$ is the standard contact form in $\mathbb{R}^{3}$ and $g_{\mathcal{H}_0}$ is a positive definite metric in  the horizontal distribution $\mathcal{H}_0$. Let $u:\Omega\subset {\mathbb{R}}^{2}\rightarrow {\mathbb{R}}$ be a ${C}^2$ function.
We denote by $g=(g_{ij})_{i,j=1, 2}$ the metric matrix and by $b=g^{-1}=(g^{ij})_{i,j=1,2}$ the inverse metric matrix. Then the mean curvature of the graph $G_u$, computed with respect to the downward pointing normal, is given by
\begin{equation}
\label{eq:hgu}
-\divv\bigg(\frac{b(\nabla u+F)}{\escpr{\nabla u+F,b(\nabla u+F)}^{1/2}}\bigg)+\mu,
\end{equation}
where $\mu$ is a bounded function in $\Omega\setminus\Omega_0$, and $\divv$ is the usual Euclidean divergence in $\Omega$.
\end{lemma} 

Furthermore in dimension three we get

\begin{lemma}\label{equivcurvaturamedia}
Let us consider the contact sub-Riemannian manifold $(\mathbb{R}^{3},g_{\mathcal{H}_0},\omega_0)$, where $\omega_0$ is the standard contact form in $\mathbb{R}^{3}$ and $g_{\mathcal{H}_0}$ is a positive definite metric in  the horizontal distribution $\mathcal{H}_0$. Let $u:\Omega\subset {\mathbb{R}}^{2}\rightarrow {\mathbb{R}}$ be a ${C}^2$ function. Then
\[
\divv\bigg(\frac{b(\nabla u+F)}{\escpr{\nabla u+F,b(\nabla u+F)}^{1/2}}\bigg)=det(g)\divv\bigg(\frac{\nabla u+F}{\escpr{\nabla u+F,\nabla u+F}^{1/2}}\bigg)+\rho,
\]
where $\rho$ is a bounded function in $\Omega\setminus\Omega_0$.
\end{lemma}

\begin{proof}
The proof is a standard computation. We only note that $\rho$ is of the form 
\[
\frac{\rho_1(b)(u_x-y)^3+\rho_2(b)(u_y+x)^3+\rho_3(b)(u_x-y)^2(u_y+x)+\rho_4(b)(u_x-y)(u_y+x)^2}{(g^{11}(u_x-y)^2+2g^{12}(u_x-y)(u_y+x)+g^{22}(u_y+x)^2)^{3/2}},
\]
where $\rho_i(b)$ are sums and products of the coefficients $g^{ij}$.
\end{proof}


Let $\Sigma\subset M$ a $\mathcal{C}^2$ surface and let $p\in\Sigma_0$. Then there exists a neighborhood $U$ of $p$ that is a Darboux chart and $\Sigma$ can be view as a graph $G_u$ in $({\mathbb{R}}^{3}, g_{\mathcal{H}_0},\omega_0)$ above defined. The projection $\Omega