Assume $\{e_1,...,e_{n+1}\}$ is the standard basis of $\mathbb R^{n+1}$. Then, $L=\{x\in \mathbb R^{n+1} : |x-\langle x, e_1 \rangle e_1 |=1 \}$ is n-cylinder of radius 1. Consider a smooth function $f:L\rightarrow \mathbb R$ such that $f(x)>-1 , \forall x\in L$. Then we have a cylinderical graph of $f$ i.e. $$ G_f=\{(x_1,(1+f)x_2,...,(1+f)x_{n+1})\in \mathbb R^{n+1} : \forall (x_1,...,x_{n+1}) \in L\} $$ Then, how to calculate the normal vector of $G_f$ ?
What I try :
$\forall x \in L$, $x_{n+1}$ can be written as $\sqrt{1-\sum\limits_{i=2}^nx_i^2}$. So $L$ can be presented as $$ \{(x_1,x_2,....,x_n,\sqrt{1-\sum\limits_{i=2}^nx_i^2} ~)\in \mathbb R^{n+1}: 1-\sum\limits_{i=2}^nx_i^2\ge 0\} $$ Equivalently, $(x_1,...,x_n)$ is cylindrical coordinates of $L$. Then the graph $G_f$ can be treat as $$ u:(x_1,...,x_n)\rightarrow (x_1,(1+f)x_2,...,(1+f)\sqrt{1-\sum\limits_{i=2}^nx_i^2}) $$ Then $\{\frac{\partial u}{\partial x_i}\}_{i=1}^n$ are basis of tangent space of $L$. Theoretically, I can get the normal vector, but it is hard to calculate and has complex form.
Are there any simple expression liking the normal vector of the graph of function on plane i.e. $\nu=\frac{(-\nabla f , 1)}{\sqrt{1+|\nabla f |^2}}$ ?