= Solution
Because the profile is parametrized by arc length,
$$
Y'^2+Z'^2=1.
$$
The tangent vectors are
$$
\phi_u=(Y'\cos v,Y'\sin v,Z'),
\qquad
\phi_v=(-Y\sin v,Y\cos v,0),
$$
so the <first fundamental form of an arc-length surface of revolution> is
$$
I=du^2+Y(u)^2\,dv^2.
$$
With the orientation determined by $\phi_u\times\phi_v$, the Gauss map is
$$
N=(-Z'\cos v,-Z'\sin v,Y').
$$
The second derivatives give
$$
e=N\cdot\phi_{uu}=Y'Z''-Z'Y'',\qquad
f=0,\qquad
g=N\cdot\phi_{vv}=YZ',
$$
and hence
$$
II=(Y'Z''-Z'Y'')\,du^2+YZ'\,dv^2.
$$
Differentiating $Y'^2+Z'^2=1$ yields $Y'Y''+Z'Z''=0$, from which
$$
(Y'Z''-Z'Y'')Z'=-Y''.
$$
Therefore
$$
\boxed{K=\frac{eg}{EG}=-\frac{Y''}{Y}},
$$
while
$$
\boxed{
H=\frac12\left(Y'Z''-Z'Y''+\frac{Z'}Y\right)}
$$
for the chosen orientation.
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