Showing that an equation of a curve in the plane defines a surface in $R^3$. A generalized cylinder is a ruled surface for which teh rulings are all Euclidean parallel. Thus there is always a parametrization of the form 
$$\mathbf{x}(u,v)=\beta (u)+v\mathbf{q} \;  (\mathbf{q}\in \mathbb{R^3}).$$
Prove: If $C:f(x,y)=a$ is a Curve in the plane, show that in $\mathbb{R^3}$ the same equation defines a surface $\mathfrak{C}$. If $t\to (x(t),y(t))$ is a parametrization of $C$, find a parametrization of $\mathfrak{C}$ that shows it is a generalized cylinder.
I'm having difficulty proving this statement. Since the equation only contains two variables, I don't know how to extend this to $R^3$ space. I would greatly appreciate some help.
 A: The curve $C$ is defined by a level-set function $f(x,y)$ as 
$$
C := \big\{(x,y)\in \mathbb{R}^2\, \big|\ \  f(x,y) = a \big\}.
$$

Note that the function $f:\mathbb{R}^2 \to \mathbb{R}$ is defined on a two-dimensional domain; therefore you cannot use this function for describing anything in $\mathbb{R}^3\!$. However, you were asked to define a surface using the same equation. So, technically what you do is you define an extension of $f$ in $\mathbb{R}^3\!$. Very often people abuse notations and use the same symbol for the  extension and for original function. For the sake of clarity, let us avoid this common practice here.

Define the extension  $F: \mathbb{R}^3 \to \mathbb{R}$ of $f$ into $\mathbb{R}^3$. This extension can be defined as $ F(x,y,z) : = f(x,y)$. Then, define generalized cylinder as the following set:
$$
\mathfrak{C} := \big\{(x,y,z)\in \mathbb{R}^3\, \big|\ \  F(x,y,z) = a \big\}.
$$
Using definition of $F$, we can write
$$
\mathfrak{C} := \big\{(x,y,z)\in \mathbb{R}^3\, \big|\ \  f(x,y) = a \big\} .
$$
But then
$$
\mathfrak{C} = \big\{(x,y)\in \mathbb{R}^2\, \big|\ \  f(x,y) = a \big\}
\times  \big\{z\in \mathbb{R} \big\} = C \times P,\tag{*}\label{0}
$$
where $P := \{z\in \mathbb{R} \} $ and $\times$ stands for Cartesian product. 
Observe that


*

*$C\in\mathbb{R}^2 $ is a plane curve, i.e. a $1$-dimentional smooth manifold embedded into $\mathbb{R}^2\!$, 

*$P \sim \mathbb{R}$ is a straight line, which is also a $1$-dimentional smooth manifold, we have 

*$C \not\subseteq P$ and $C \not\supseteq P$  because $P$ is orthogonal to the plane of $C$. 


In this way, our cylinder $\mathfrak{C}$ is defined as a Cartesian product of smooth manifolds of dimensionality $1$. Therefore $\mathfrak{C}$ is itself a smooth manifold of dimension $2$ embedded into $\mathbb{R}^3\!$!
Now we only have to derive parametrization of $\mathfrak{C}$. 
First, recall that any  plane curve $\gamma$ can parametrized in $\mathbb{R}^2$ by a smooth map
$$
\gamma : I \to \mathbb{R}^2,
$$
where $I\in\mathbb{R}$ is some interval ( a non-empty connected subset of $\mathbb{R}$) and $\gamma(t) =  \big[\,x(t),\, y(t)\, \big]'$. 
Since we need the parametrization in $\mathbb{R}^3$, we can use extension of $\gamma$ and define a new parametrization
$$
\beta(t) = 
\begin{bmatrix}
x(t)\\ y(t) \\ 0
\end{bmatrix}
: I \to \mathbb{R}^3, \qquad \text{ where } \quad
\gamma(t) = 
\begin{bmatrix}
x(t)\\ y(t) 
\end{bmatrix}\tag{1}\label{1}
$$
Second, denote $\vec{n} = [0,0,1]$ — the unit normal vector for domain plane of $C$. 
Then $P$ can be parametrized in $\mathbb{R}^3$ by the map $\mathbf{v}: \mathbb{R} \to \mathbb{R}^3$
$$\tag{2}\label{2}
\alpha(s) = s\,\vec{\mathbf{n}} = 
\begin{bmatrix}
0 \\ 0 \\  s
\end{bmatrix},
\quad s \in \mathbb{R}
$$
Finally, combining parameterizations $\eqref{1}$ and $\eqref{2}$ in the Cartesian product expression $\eqref{0}$ for $\mathfrak{C}$, we get the following parametrization for generalized cylinders:
$$
\mathbf{r}(t,s) = \beta(t) + \alpha(s) = \beta(t) + s\,\vec{\mathbf{n}}, 
$$
where $t\in I\subset\mathbb{R}$, $s\in\mathbb{R}$, $\beta(t)= \big[\,x(t),\, y(t),\, 0\,\big]'$, and $\vec{\mathbf{n}} = [0,0,1]'$
