Determine the result of parallel translating the vector $(0, 0, 1)$ once around the circle $x^2+ y^2=a^2, z=0$ can any one solve this problem 

Q)
  Determine the result of parallel translating the vector $(0, 0, 1)$ once around the circle $x^2+ y^2=a^2, z=0$ .On the right circular cylinder $x^2 + y^2 =a^2$?

what happen if 
Determine the result of parallel translating the vector $(0, 0, 1)$ once around the circle $x^2+ y^2=a^2, z=0$ .On the right circular cone $x^2 + y^2 =a^2$?
I hope someone can solve this problem and answer also what is the parametrization of cone and cylinder in this case.
Thanks. 
 A: Some parts of the question are not very clear, so I'll just address the part that says: "determine the result of parallel translating the vector $(0,0,1)$  once around the circle $x^2 + y^ 2 = a^2$, $z=0$".
Since this is an elementary question, I expect "the vector $(0,0,1)$" actually means a line in the direction $(0,0,1)$. This is the thick blue line in the picture. Let's orient ourselves so that this line is vertical. Then the given circle lies in a horizonal plane. This circle is shown in red in the picture. If the blue line travels around the red circle, it will sweep out a cylindrical shape, shown shaded in pale blue. In fact, this cylinder will have the equation $x^2 + y^ 2 = a^2$. This is a right circular cylinder with radius $a$ (the same as the circle).

I already answered your questions about parameterizations of cylinders and cones, but I'll repeat part of my answer here.
The cylinder we have been discussing has a parameterization:
$$P(u,v) = (a\cos u, a\sin u, v)$$
It is helpful to write this as
$$P(u,v) = (a\cos u, a\sin u, 0) + v(0,0,1)$$
Let's think about what this says. Given two parameter values $u$ and $v$, this equation tells us how to get to a point $P(u,v)$ on the cylinder. The picture below shows how this works.

The equation says we should first go to the point $(a\cos u, a\sin u, 0)$, which at an angle $u$ on the red circle. This is the green point in the picture. Then we should travel vertically a distance $v$ along the vector $(0,0,1)$ (i.e. up one of the blue lines). This brings us to the pink point in the picture, which is $P(u,v)$.
From this, you should be able to see how the blue lines generate the cylinder.
A: The cylinder is a flat space (it is a curled up Euclidean plane) so the result is $(0,0,1)$.
