Let $ u : \Omega\to \mathbb R$, where $\Omega$ is an open set in $\mathbb R^2$. Let $a\ge 0$. Consider the following integral:
$$ \int_\Omega \frac{\det D^2 u}{(1+ |D u|^2)^a} \mathrm d x \mathrm d y.$$
In an answer to another question, it is shown that the above integral depends only on the value of $Du, D^2u$ on the boundary $\partial \Omega$. This is done using some simple argument in calculus of variations. This result is suggesting the following
Question: Can the above integrand be (explicitly) written as divergence of something?
When $a=0$ it is easy:
$$\int_\Omega \det D^2 u \mathrm dx \mathrm dy = \int_\Omega (u_{xx}u_{yy} - u_{xy}^2 )\mathrm dx\mathrm dy = \int_\Omega \mathrm{div}\big( u_xu_{yy} , -u_x u_{yx}\big) \mathrm dx\mathrm dy.$$
When $a = 3/2$, the above integration can be written as $$ \int_\Omega K \mathrm dA,$$ where $K$, $\mathrm dA$ are respectively the Gaussian curvature and the area element of the graph $(x, y, u(x, y))$. Thus, again, the integrand can be written as $K\mathrm dA= \mathrm d\omega_{12}$, where $\omega_{12}$ is basically the Christoffel symbols given by
$$ \nabla e_1 = \omega_{12} e_2,$$
where $\{e_1, e_2\}$ is an orthonormal frame on the surface (the graph). So $\{e_1, e_2\}$ and thus $\omega_{12}$ can be represented as partial derivatives of $u$ so it is sort of explicit. (Remark: in general for a surface, if the first fundamental form is diagonal $(F=0$), then $$K \mathrm dA= \mathrm{div}\bigg( \frac{\sqrt g}{E} \Gamma_{xx}^y \ , - \frac{\sqrt g}{G} \Gamma_{xy}^y\bigg) \mathrm dx\mathrm dy,$$ which is divergence of something. But for a graph $(x, y, u(x, y))$, the first fundamental form is in general not diagonal and it is extremely messy to write down explicitly, though doable).
Thus the answer to the question is yes at least when $a = 0, 3/2$. What about the general situation?
Also, if the answer is yes, are there any geometric interpretations? (I guess it would be no, since the curvature should be the only geometric invariance).