# Computing a discrete surface integral of a scalar function

1. Oct 24, 2013

### 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

\label{eq:parametrization}
\mathbf{v}(s,t) = (1-t)\left[ (1-s)\mathbf{v}_k + s\mathbf{v}_j \right] + t \mathbf{v}_i,

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

\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).

Hence, the surface integral can be transformed into the following double integral:

\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 },

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}