# Net flow into a tetrahedron?

1. Jun 21, 2010

### TopCat

I'm working through Advanced Calculus: A Differential Forms Approach at my leisure. In going over two-forms the notions of "flow across an area" and "oriented area" are introduced I hit a brick wall with grasping orientations, though, when asked to find the total flow into a tetrahedron.

So this is what's confusing me. For a unit flow in the z-direction of xyz-space, the flow across a surface is just the oriented area dxdy of the surface. So all I have to do to answer the question is find the oriented areas of the projections of all four component triangles of the tetrahedron on the xy-plane and add them together. The result should clearly be zero since, intuitively, the flow is uniform so that nothing is accumulating inside the tetrahedron and, therefore, flow in = flow out.

If I have a regular tetrahedron with vertices PQRS, PQR in the xy-plane and S at point (0,0,1), then the projection looks like a triangle with S at the centroid.
[PLAIN]http://www.j1n.org/srs/misc/tetra.JPG [Broken]
Since each triangle is oriented in the direction $$P \rightarrow Q \rightarrow R \rightarrow S$$ then we have the four component oriented triangles PQR, PQS, QRS, and PRS.

Since $$A_{PQS} + A_{QRS} + A_{PRS} = A_{PQR}$$ (where A denotes area), each of the smaller triangles must have an orientation opposite the big triangle so the total sum of oriented areas is zero. But given the orientation of the vertices, QRS, PQR and PQS have the same orientation (all are counter-clockwise) so the total flow into won't sum to zero.

So clearly I've oriented two of the triangles wrong. I'm confused as to how I should establish orientation if I can't go by the handedness of the direction given by the vertices.

Last edited by a moderator: May 4, 2017
2. Jun 21, 2010

### LCKurtz

Clockwiseness is in the eye of the beholder. Imagine yourself moving around the tetrahedron on the outside so you can look directly at each face. In your picture you already are pretty much viewing face PQS and QRS and they are counterclockwise from your viewpoint, which would correspond to an outward normal by the right hand rule.

If you are standing on the left of the object, you would have to use PSR for the same counterclockwise orientation, and if you were underneath looking up you would use PSQ.

Also you wouldn't expect zero total flow unless the flow was incompressible, for then by the divergence theorem

$$\int\int_S \vec F \cdot d\vec S = \int\int\int_V \nabla \cdot \vec F\, dV= 0$$