Computing a discrete surface integral of a scalar function

[teodron]

Consider a triangulated discrete manifold (a polyhedron) with known vertices (i.e. each vertex is given in terms of its $$(x,y,z)$$ coordinates ).

Assign scalar values (some kind of potentials) to each vertex (i.e. at each vertex, a $$k_t(\mathbf{v})$$ is known through its value, no analytical expression is given!).

For any triangular face of this discrete manifold, one is required to compute the surface integral of the $$k_t$$ scalar discrete field:
$$ \int_{\Delta}{k_t dS} $$

I have a small sketch for a solution, but I do not feel it is mathematically sound. Could anyone follow it briefly and see what goes wrong (if anything?)?

Let
\begin{equation}
\label{eq:parametrization}
\mathbf{v}(s,t) = (1-t)\left[ (1-s)\mathbf{v}_k + s\mathbf{v}_j \right] + t \mathbf{v}_i,
\end{equation}
with $ (s,t) \in [0,1]^2 $ be a parametrization for $ \Delta $ . If $f \equiv k_t $ is the energy function defined over a surface patch, then
\begin{equation}
\label{eq:energyParametrization}
f(\mathbf{v}(s,t)) = (1-t)[(1-s) f(\mathbf{v}_k) + s f(\mathbf{v}_j)] + t f(\mathbf{v}_i).
\end{equation}
Hence, the surface integral can be transformed into the following double integral:
\begin{equation}
\label{eq:energySurfaceIntegral}
\int_{\Delta_{\mathbf{v}_i\mathbf{v}_j\mathbf{v}_k}} {f(x,y,z) dS} = \int_0^1 \int_0^1 {g(s,t) \left\lVert \frac{\partial \mathbf{v}}{\partial s} \times \frac{\partial \mathbf{v}}{\partial t} \right\rVert ds dt },
\end{equation}
where $$g(s,t) = f(\mathbf{v}(s,t)) $$. This last expression is directly computable and leads to the following discrete bending energy
formulation over the 1-ring of a $$\mathbf{v}_i$$ vertex:
\begin{multline}
\label{eq:bendingEnergyDiscrete}
E_{bending}(\mathbf{v}_i) = \sum_{\Delta_{\mathbf{v}_i\mathbf{v}_j\mathbf{v}_k} \in \mathcal{N}^1(\mathbf{v}_i)}
\frac{3f(\mathbf{v}_j) + 3f(\mathbf{v}_k) + 4f(\mathbf{v}_i)}{15} \cdot \\
\cdot \left\lVert \mathbf{v}_j\times\mathbf{v}_i - \mathbf{v}_j\times \mathbf{v}_k - \mathbf{v}_k\times\mathbf{v}_i \right\rVert .
\end{multline}

 

[pat_connell]

This method is wrong because it assumes that the function $f$ is linear, which may not be the case in general. In order to compute the integral correctly, one should use the full expression of the function $f$ and integrate it over the triangular face.