Let $A= \{(0,y,z):z^2 + (y-2)^2=1 \}$ and let $B$ be the set obtained by rotation of $A$ around the $z$-axis. Determine a parametrization for $B$. I am stuck on the following problem:

Let $A= \{(0,y,z):z^2 + (y-2)^2=1 \}$ and let $B$ be the set obtained by rotation of $A$ around the $z$-axis. Determine a parametrization for $B$.

I have a faint idea that the parametrization should be along the lines of:
\begin{align}x&= 2+\cos u \\y&=2 + \cos v \\ z&=\sin v \end{align}
In which $y,z$ are the polar coordinates for the circle given in $A$ in the $y,z-$plane. As $B$ is the rotation along the $z$-axis, I guessed $x$ should be something like $x=2 + \cos \psi$ akin to $y$, but the answer in the book is:
\begin{align}x&=(2+ \cos v) \cos u \\ y&=(2 + \cos v) \sin u \\ z&=\sin v\end{align}
I checked on Wolfram Alpha and my parametrization describes a tube while the solution (correctly) describes a torus, but I don't know how to arrive at it. Is there some general procedure? The book I am reading basically said: "Look at these (simple) examples. Now handle this". 
I think that taking the vector $(2+\cos v,2+ \cos v )$ we'd need to "regulate" it's length by multiplying each coordinate by $\cos u, \sin u$ but I can't put it into words in a way that makes sense. Perhaps thinking about polar coordinates with $r=2+\cos v$?
 A: Originally, you write $z$ and $y-2$ as the sine and cosine of some angle. So in step 1, $z_0=\sin v$ and $y_0-2=\cos v$. Since you are rotating around the $z$ axis, whatever $y_0$ you have originally will transform into $(x,y)=(y_0\sin u,y_0\cos u)$. Putting it all together: $$x=y_0\sin u=(2+\cos v)\sin u\\y=y_0\cos u= (2+\cos v)\cos u\\z=z_0 = \sin v$$
A: I "discovered" it. We just need to parametrize the curve in the $y,z$ plane and apply the rotation matrix 
$$\small \begin{bmatrix}
\cos \theta &  -\sin \theta & 0 \\[3pt]
\sin \theta &   \cos \theta & 0\\[3pt]
0 & 0 & 1\\
\end{bmatrix} \begin{bmatrix}
0 \\
2+\cos(v) \\
\sin(v) \\
\end{bmatrix} = \begin{bmatrix}
-\sin(\theta)(2+\cos (v)) \\
\cos(\theta)(2+\cos(v)) \\
\sin(v) \\
\end{bmatrix}$$
which yields the desired result with the exception of the discrepancy of $x$ but the surface in question is also the torus. Perhaps the orientation is changed?
I guess the method is this (at least for most of the exercises I've met): You get the coordinates in a plane and apply some rotation matrix on them. 
This is probably what Andrei tried to tell me (without mentioning the matrix) but I was too dumb to understand. 
