What is the boundary of the surface area of a cone?

Question

To verify Stokes' theorem on a cone, I need to calculate the line integral of the given vector field around the 1 dimensional boundary of the surface of the cone. However, I can't imagine what the boundary would be in this case, since the surface seems to have none. What is the boundary of the surface area of a cone?

Answer 1

Stokes' theorem can only be applied to compact surfaces and their boundaries. In order to do an example, either a certain finite cone is given to you, or you have to invent one, say $$C:=\bigl\{(x,y,z)\bigm| 0\leq z=\sqrt{x^2+y^2}\leq1\bigr\}\ .$$ This $C$ is just given as a point set. Orient it such that the normal $n$ points outwards.

This cone has a peak, where it is not smooth. But this is no problem. Just make sure that you have a clean parametrization of $C$ giving the normal you want.

The boundary curve $\partial C$ is the unit circle at level $z=1$. The sense of direction of $\partial C$ has to be chosen such that it corresponds with the chosen $n$: When you look from the arrow head of $n$ "down" to $C$ the curve $\partial C$ has to go counterclockwise. In terms of the $(x,y,z)$ coordinate system this means that $\partial C$ goes clockwise when seen from $z=\infty$.