I want to show that the sets $$S=\{(x,y,z)|0<x<a,z>0,y^2+z^2=b^2\}$$ and $$C=\{(x,y,z)|x>0,y>0,z>0,z^2=1-x^2-y^2\}$$ are non-singular open $C^2$-surfaces.
The only definition I know about this question is "A singular $C^m$-surface of dimension $k$ in $\mathbb{R}^N$ is $S=(U,\phi)$, where $U\subset\mathbb{R}^k$, $\phi:U\to\mathbb{R}^N,\phi\in C^m.$"
I have no idea about this problem. Anyone can give me a hint or a better definition? Thanks
The definition you gave is just for "surface". What makes the surface singular is if the rank of $\phi$ is ever less than $k$.
For $S$, we can take $\phi$ to be $\phi\ :\ (x, y) \mapsto (x, y, \sqrt{b^2 - y^2})$.
For $C$, we can take $\phi$ to be $\phi\ :\ (x, y) \mapsto (x, y, \sqrt{1 - x^2 -y^2})$.
To show that $\phi$ is non-singular, show that $D_{(x,y)}\phi$ is of rank $2$ for all $(x, y)$ in the domain. That should not be too hard to do.
By open, I presume it means that the surfaces do not include the borders. That can be shown from the limitations of $x, y, z$, which are all expressed as strict inequalities.