Show that this is indeed a differentiable manifold with boundary. I want to show that the cylinder:
$$C = \{ (x,y,z)\in\mathbb{R}^3: x^2 + y^2 = 1, 0 \le z \le 1 \}$$ 
is indeed a a differentiable manifold with boundary, this means the following:
A subset $M \subset \mathbb{R}^n$ is a differentiable manifold of dimension $k$ with boundary if, for each $x \in M$, exists open sets $U \subset \mathbb{R}^k$ and $V \subset \mathbb{R}^n$ and a class $C^1$ function $f:U \to V$ such that:
1) $x \in V$ 
2) $f:U\cap \mathbb{H}^k \to V\cap M$ is an homeomorfism
3) for each $y \in U\cap \mathbb{H}^k $ the jacobian matrix has rank k
and $\mathbb{H}^k=\{x\in \mathbb{R}^k : \text{the k-th component is not zero}\}$ 
So I was thinking in $f$ as the function $f(x,y)=(x \sqrt{1-x^2},y$ but I don't know how to take $U$ and $V$ such that the conditions are satisfied (I was thinking $U=[0,1]$ and $V=(0,1]$ but I don't know if it may help).
Thanks a lot for your help in advance 
My way of writing the proof:
I take the function as in the answer, since my definition requires the existence of two open sets $U,V$ then I choose $U=(0,2\pi)\times (0,1)$ then we note that $U \cap \mathbb{H}^2=U$, Now, my question is how to take the optimum $V$ for the definition requirements, because if I take $V=\{(x,y,z)\in R^3:|x|<1,|y|<1,0<z<1 \}$ the intersection $V\cap C$ is not what we need for the definition

Take the parametrization 
$$ f(\theta,z) = (\cos \theta, \sin \theta, z ) $$
From here you have your Jacobian; clearly it always has full rank. 
Edit: For any interior point $x \in M$ on the cylinder, we have that
$$U = (0, 2\pi) \times (0,1) \quad \& \quad V= \{ x,y,z:x^2+y^2=1 , z \in (0,1)\} $$
or
$$U = (- \pi, \pi) \times (0,1) \quad \& \quad V= \{ x,y,z:x^2+y^2=1 , z \in (0,1)\} $$
will suffice. For $x \in M$ on the boundary, we have the half balls 
$$U = (0, 2\pi) \times [0,1) \quad \& \quad V= \{ x,y,z:x^2+y^2=1 , z \in [0,1)\} $$
or
$$U = (- \pi, \pi) \times [0,1) \quad \& \quad V= \{ x,y,z:x^2+y^2=1 , z \in [0,1)\} $$
for the bottom, and
$$U = (0, 2\pi) \times (0,1] \quad \& \quad V= \{ x,y,z:x^2+y^2=1 , z \in (0,1]\} $$
or
$$U = (- \pi, \pi) \times (0,1] \quad \& \quad V= \{ x,y,z:x^2+y^2=1 , z \in (0,1]\} $$
for the top.
