In the article: <https://arxiv.org/abs/0906.3217> the authors prove in Lemma 1 a formula which helps compute more easily the integral of the Hessian of a function defined on $\Bbb{S}^2$. More precisely, if $h : \Bbb{S}^2 \to \Bbb{R}$ is a $C^2$ function, $Hess(h)(X,Y) = \langle \nabla\_X \nabla h,Y\rangle$ is the Hessian of $H$ and **$H(h)$ is the determinant of the Hessian**, then 
$$ \int\_{\Bbb{S}^2} H(h) dA = \frac{1}{2} \int\_{\Bbb{S}^2} |\nabla h|^2 dA$$
where $dA$ is the area element on $\Bbb{S}^2$. This formula is helpful and simplifies some aspects of my numerical computations. Here and in the following, $\nabla$ and $\Delta$ represent the tangential gradient and Laplace-Beltrami operator on $\Bbb{S}^2$.


The above formula is proved in connection with bodies of constant width, but in the proof in the article they don't seem to use this fact. In my numerical experiments the formula gives the expected result in the general case. However, I work in the case where $H(h)>0$ (I don't know if this is relevant or not...)



> 
> I was wondering if it is possible to obtain a similar simplification for the integral $\int\_{\Bbb{S}^2} h H(h) dA$? More precisely, is it possible to obtain something of the form
>  $$ \int\_{\Bbb{S}^2} h H(h) dA = \int\_{\Bbb{S}^2} \mathcal{F}(h,\nabla h,\Delta h)dA $$
>  where $\mathcal{F}$ is "simple" (polynomial)?
> 
> 
> 



Simplification of integral on the sphere
There *is* a formula that is roughly of the kind that the OP desires. On a closed, oriented Riemannian surface $(M,g)$ with Gauss curvature $K$, the formula is
$$
\int\_MuH(u)\,\mathrm{d}A = \frac12\int\_M \left(K u |\nabla u|^2 
- \nabla^2(u)(J\nabla u,J\nabla u)\right)\,\mathrm{d}A.
$$
Here, $\nabla^2(u)$ is what the OP called the Hessian quadratic form of $u$ and $J:TM\to TM$ is the `rotation by $\pi/2$ operator' on tangent vectors.


I do not know whether the right hand side can be further simplified so that the polynomial depends only on $u$, $\nabla u$, and $\Delta u$.


