Prove: The boundary of the set $\{(x,y,z) \in \mathbb{R}^3 | z > \sqrt{x^2+y^2}\}$ isn't a smooth manifold.
The boundary is defined by $z = \sqrt{x^2+y^2}$. I'm trying to think how to approach this problem. I know that the problematic point is $0$, because intuitively, it seems like the mapping isn't smooth at that point.I know that because smooth manifolds are locally homeomorphic to $\mathbb{R}^n$, so in our case the the manifold around the point (0,0,0) has to be homeomorphic to either an open set in $\mathbb{R}$ or in $\mathbb{R^2}$. The former can't be true because removing one point from the manifold, for example, (0,0,0), doesn't disconnect it but removing one point from $\mathbb{R}$ does.
I'm stuck because I can't see a way to topographically disprove the second case. I though about another approach: Assume to the contrary that the boundary is a manifold, therefore there exists some co-map $f$ of the manifold at point $(0,0,0)$ s.t.: $f(x,y, \sqrt{x^2+y^2}) = 0$. I tried deriving both sides, but didn't get anywhere. It seems like I'm out of ideas to disprove the question... Any help would be welcome.