# How to get Gauss points for nodal force vector (surface integral) in tetrahedral elements in the isoparametric coordinate system $({\xi \eta \zeta})$?

I am aware of the following question Evaluating the surface integral in an FEM (Finite Elements Method) procedure. But they use the volumetric coordinates while I want to use the cartesian isoparametric coordinates.

I am using the cartesian coordinate system ($$xyz$$) and the corresponding isoparametric coordinate system $$({\xi \eta \zeta})$$. Then the elemental force vector is given by

$$\textbf{f}^e = \int_{\partial{V_{xyz}}} \textbf{N}_i \textbf{t}_n d({\partial{V_{xyz}}})$$

$$= \int_{\partial{V_{\xi \eta \zeta}}} \textbf{N}_i \textbf{t}_n \Vert \textbf{J} \Vert d({\partial{V_{\xi \eta \zeta}}})$$

$$= \sum_{i=1}^{\#Gauss.points} \textbf{N}_i \textbf{t}_n \Vert \textbf{J} \Vert W_i$$

Assuming a generalized case where the cartesian element may even be inclined i.e., none of the coordinate valued need to be '0'. My confusion here is that since the integral is over a surface of the tetrahedron, how do I evaluate the Jacobian of the surface? I thought the best way in such a case is to use the 3D shape functions directly. Now the 3D shape function vectors $$N$$ contain all coordinates ($$\xi \eta \zeta$$) or I cannot say which will be '0', while the iso-parametric triangle for surface is a 2D triangle with only coordinates ($$\xi \eta$$). So I would ideally need surface integration points in an ($$\xi \eta \zeta$$) system? How do I go about this? I see that all FEM notes trivialize this. What am I misunderstanding then?

• How do you define your "cartesian isoparametric coordinate" system? What are the shape functions for a tetrahedral element in this system? And how do you perform the volume integration? I do not see how such a system is particularly useful. Jan 9 at 18:34
• by cartesian isoparametric system i only refer to the traditional isoparametric system used in finite element (with lagrange shape functions). I just wanted to differentiate it from the volumetric coordinates used 'only' in triangles and tetrahedrons. it is useful because i could then generalize it for hexahedral elements unlike the volumetric coordinates. Jan 9 at 21:55