How to understand the integral? In order to understander the nonlinear elliptic equation with natural boundary condition,
$$\sigma_2(D^2u)=0 \text{ in } \Omega$$
I wish to understand the following integral,
$$E(u,\Sigma)=\int_\Sigma \det(D^2u|_\Sigma) \, de_1\wedge de_2$$
Where $u\in C^2(\Omega)$ is a solution of some elliptic equation, so we have $D^2u$ is semi-positive, $\Omega\subset R^n$ is a open set, $\Sigma$ is a arbitrary smooth surface equipped with the induce metric from $R^n$, and $\Sigma\subset \Omega$, $\{e_1,e_2\}$ is a pair of orthogonal biases of $\Sigma$.
I wish to understand if there is a similar result like Newton-Leibniz formula in the one dimensional case, which is the following:
$$\int_{\gamma}\partial_{e_1e_1} u \, de_1=\int_{\partial \gamma} u = \partial_{e_1} u(\gamma(1))-\partial_{e_1}u(\gamma(0))$$
Where $\gamma: [0,1]\to \Omega$ is a $C^2$ curve and $e_1$ is the gradient direction along the curve.
So my problem is following:

Problem
  Can we get some integral expression for $E(u,\Sigma)=\int_\Sigma \det(D^2u|_\Sigma) \, de_1\wedge de_2$ use just $u_1=\partial_{e_1}u,u_2=\partial_{e_2}u $, and the integral domain of the expression is $\partial \Sigma$? more precisely I wish we could find a functional $\hat E(u,u_1,u_2)$ such that,
  $$E(u,\Sigma)=\int_\Sigma \det(D^2u|_{\Sigma}) \, de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)?$$
  $\hat E$ is a functional only relate to $u,\nabla u$. Because of $\hat E$ only related with $u,\nabla u$, I would like to say it is the lift of $E(u,\Sigma)$. And may be this is only true for some special domain $\Sigma$, for example $\Sigma$ is a Ball, this is exactly what I excepted.

Motivation
Now I need to explain my motivation why I except this is true and why I need this or it variation is true.
The motivation is come from the 1 dimensional version is true, and which is crucial to establish the mean value principle for Laplace equation,
$$\frac{1}{\mu(B)}\int_{\partial B(r,x_0)} u(x) \, dx=u(x_0)$$
I wish to generated the mean value principle to some special nonlinear elliptic equation by this way, although this mean value principle may be not exist, I still wish to explain why the mean value property failed by this way.
Attempt
I have four ways to attempt this problem, but there always emerge difficulties I could not settle.


*

*Use the identity 
$$\exp(\operatorname{tr}(A))=\det(\exp(A))$$
We try to solve the equation $D^2 u=\exp\operatorname{tr} (A) \tag{$*$},$ we pretend it could be solved then we have: $A=\log(D^2(u))=\sum_{k=0}^\infty \frac{(-1)^{k+1}}{k} {D^2(u)}^k$, so we have,
$$E(u,\Sigma)=\int_\Sigma e^{\operatorname{tr}\left(\sum_{k=0}^\infty \frac{(-1)^{k+1}}{k}{D^2(u)}^k\right)} \, de_1\wedge de_2$$
It seems much easy to find $\hat E(u)$ use stokes theorem with $E(u)$ under this form.
But there exists two problem, one is that the solvable of $(*)$ and there exists infinity many different solution of $(*)$ if my insight is right and I do not know how to proof the identity we could proved in this way is independent with the choice.


2.
We discretization the problem and consider it in $\mathbb Z^2$ which is a two dimensional affine subspace of $\mathbb Z^n$.
The advantage of discretization is that we can explicate calculate $u_{11},u_{12},u_{21},u_{22}$ now, in fact,
$$u_{11}(x,y) = h^2(u(x+2h,y)+u(x,y)-u(x+h,y)-u(x+h,y))        $$
$$u_{12}(x,y) = h^2(u(x+h,y+h)+u(x,y)-u(x+h,y)-u(x,y+h))       $$
$$u_{21}(x,y) = h^2(u(x+h,y+h)+u(x,y)-u(x+h,y)-u(x,y+h))       $$
$$u_{22}(x,y) = h^2(u(x,y+2h)+u(x,y)-u(x,y+h)-u(x,y+h))        $$
and we could use this to calculate $\det(D^2u)$, but after calculate I do not find general principle and what shape should $\Sigma$ be to make the identity,
$$\int_\Sigma \det(D^2u|_\Sigma) \, de_1\wedge de_2=\int_{\partial\Sigma}\hat E(u,u_1,u_2)$$
make sense in this way.
3.
Investigate the Frobenius integrable condition, which is just mean:
$$L_i L_j u(x)-L_jL_iu(x)=\sum_k c_{ij}^k(x)L_ku(x)$$
should be true, I tried to split $E(u,\Sigma)$ into several parts, and every part of it satisfied the Frobenius integrable condition, i.e.
$$E(u,\Sigma)=\sum_{i=1}^k E_i(u,\Sigma)$$
and $E_i$ satisfied the Frobenius integrable condition. And we investigate each $E_i$ first and combine the result we got together to establish a result for $E(u,\Sigma)$.
But the difficulties comes from that I do not know how to decompose $E(u)$ at all!
4.
The last strategy could only get a part of result(instead of identity, we could only get a inequality). thanks to the elliptic condition we know $D^2u$ is semi-positive and the Principal minors of $D^2u$ is also semi-positive, so $D^2u|_{\Sigma}$ is semi-positive. we could consider the function $f(x)=\det(D^2(u)|_\Sigma)^{1/2}$, which is a concave function so use Jensen inequality we could get following result:
$$\frac1{\operatorname{vol}\Sigma}\int_{\Sigma}(\det D^2u|_\Sigma)^{1/2} \le \det \left(\frac1{\operatorname{vol}\Sigma} \int_\Sigma D^2u|_\Sigma(x) \right)^{1/2}.$$
May be according this could gain a mean-value inequality but I am not very sure.
May be all of these approaches are useless. In any case, I wish some result could be establish, whatever positive answer or negative answer. I will appreciate to any valuable advice or new idea, thank you very much!
 A: If I understand your question correctly, the answer is affirmative for Ricci-flat (i.e. flat) surfaces (not necessarily embedded in a Euclidean space) according to equation (14) of
Reilly, Robert C., Applications of the Hessian operator in a Riemannian manifold, Indiana Univ. Math. J. 26, 459-472 (1977). ZBL0391.53019.
which is available at http://www.jstor.org/stable/24891302?seq=1#page_scan_tab_contents .  However for general surfaces there is a Ricci curvature (which is just scalar curvature on 2D) term which is unlikely to be eliminated in general (e.g. for positive curvature or negative curvature surfaces it will have a definite sign).
Added later: actually one can just rewrite the integral using covariant derivatives with the Levi-Civita connection on $\Sigma$ as $\int_\Sigma \frac{1}{2} (\nabla^\alpha \nabla^\alpha u \nabla_\beta \nabla_\beta u - \nabla^\alpha \nabla^\beta u \nabla_\alpha \nabla_\beta u)\ dg$, where $dg$ is the Riemannian volume form, and integrate by parts, picking up the curvature term thanks to the lack of commutativity of the covariant derivatives when applied to tensors.
