Let $S_1 : [0, 2\pi r]\times [0, h]$
$S_2: x^2+y^2=r^2$
Let $f: S_1 \to S_2$
$(u,v)=(r\cos (\frac{u}{r}), r\sin (\frac{u}{r}), v)$ for $v\in [0,h]$ and $u\in [0, 2\pi r)$
How do I prove that $f$ is a diffeomorphism and $f$ is an isometry?
I know that the following theorem;
$f$ isometry $\iff$ for $S_1$ and $S_2$, the surface patches have the same first fundamental form.
But I don't know how to show these. Please can you help me. Thank you.
The definition of isomorphism: let $S_1$ and $S_2$ be two regular surfaces. Let $f: S_1 \to S_2$ be diffeomorphism. $f$ is isometry if the length of $\gamma$ in $S_1$ must be equal to the length of $f\circ \gamma$ in $S_2$ for any curve $\gamma$ in $S_1$. i.e $L_{S_1}(\gamma )= L_{S_2}(f \circ \gamma )$
The definition of difeomorphsim: $f:S_1\to S_2$ is a diffeomorphism
if
(1) $f$ is smooth.
(2) $f$ is one to ne and onto.
(3) $f^{-1} :S_2\to S_1$ is smooth.