Find Christoffel Coefficients 
Evaluate the Christoffel coefficients of the following surface of revolution:
$$X(\theta,s)=(r(s)\cos\theta,r(s)\sin\theta,z(s))$$

So we first start with finding the mertic
$X_{\theta}=(-r\sin\theta,r\cos\theta,0)$
$X_{s}=(r'\cos\theta,r'\sin\theta,z')$
$g_{11}=<X_{\theta},X_{\theta}>=r^2\sin^2\theta+r^2\cos^\theta=r^2(\sin^2\theta+\cos^\theta)=r^2$
$g_{12}=g_{21}=<X_{\theta},X_{s}>-rr'\sin\theta \cos\theta+rr'\sin\theta \cos\theta=0$
$g_{22}=<X_{s},X_{s}>=(r')^2\cos^2\theta+(r')^2\sin^2\theta+(z')^2=(r')^2(\cos^2\theta+\sin^2\theta)+(z')^2=(r')^2+(z')^2$
So we get $$g_{ij}=\begin{pmatrix}r^2 & 0\\0 & (r')^2+(z')^2\end{pmatrix}$$
Now in the book I saw that $(r')^2+(z')^2=1$ why is that? can we say that all surface of revolution have the same metric?
$$g_{ij}=\begin{pmatrix}r^2 & 0\\0 & 1\end{pmatrix}?$$
Now we have to find $$g_{ij,\theta}=\begin{pmatrix}0 & 0\\0 & 0\end{pmatrix}$$
and $$g_{ij,s}= \begin{pmatrix}2r^2r' & 0\\0 & 0\end{pmatrix}$$
and $$g^{ij}=\frac{1}{r^2}\begin{pmatrix}1 & 0\\0 & r^2\end{pmatrix}$$
Now how do I calculate Christoffel coefficients?
 A: The condition $(r')^2+(z')^2=1$ does not follow from anything else; it has to be assumed separately.
This assumption is not unreasonable or restricting, though.
Consider the curve $\gamma\colon I\to(0,\infty)\times\mathbb R$ given by $\gamma(s)=(r(s),z(s))$, where $I$ is some open interval.
This curve is (hopefully) assumed to be injective — an arc.
The surface of revolution only depends on the image $\gamma(I)$ (which is then revolved around an axis), not on the paramterization.
The most convenient parametrization is often that of unit speed, so that 
$$
1
=
|\gamma'(s)|^2
=
(r')^2+(z')^2.
$$
This is precisely your condition.
You could use any parametrization whatsoever.
This would simply produce a different system of coordinates on the surface.
After all, reparametrization is nothing but a change of coordinates in dimension one.
To calculate the Christoffel symbols, you can use the formula (which you have hopefully been given):
$$
\Gamma^i_{\phantom{i}jk}
=
\frac12g^{il}(g_{lj,k}+g_{lk,j}-g_{jk,l}).
$$
You have computed all the required matrix elements (with coordinates $(x^1,x^2)=(\theta,s)$):
$$
\begin{split}
g_{11}&=r^2
\\
g_{12}&=0
\\
g_{22}&=1
\\
g^{11}&=r^{-2}
\\
g^{22}&=1
\\
g^{12}&=0
\\
g_{ij,1}&=0
\\
g_{ij,2}&=\delta_{1i}\delta_{1j}2rr'.
\end{split}
$$
(There was a typo in your $g_{11,2}$; it's $2rr'$, not $2r^2r'$.)
Now you just have to multiply and sum these up according to the formula.
You will need to calculate
$\Gamma^1_{\phantom{1}11}$,
$\Gamma^1_{\phantom{1}12}=\Gamma^1_{\phantom{1}21}$,
$\Gamma^1_{\phantom{1}22}$,
$\Gamma^2_{\phantom{2}11}$,
$\Gamma^2_{\phantom{2}12}=\Gamma^1_{\phantom{2}21}$, and
$\Gamma^2_{\phantom{2}22}$.
It may look as if all surfaces of revolution have the same metric
$$
g=\begin{pmatrix}r^2 & 0\\0 & 1\end{pmatrix}.
$$
(Notice that I denote the matrix by $g$. The matrix elements are $g_{ij}$; for example $g_{11}=r^2$. Your material may use another convention.)
However, this is not exactly true; the way the radius $r$ varies is different for different surfaces.
Remember that $r$ depends on $s$, which is one of your coordinates.
If you had two surfaces of revolution with radii $r,\tilde r$ satisfying $r(s_0)=\tilde r(s_0)$ at some $s_0\in I$, then the two metrics indeed look exactly the same at the coordinates $(\theta,s_0)$ for any $\theta$.
But this only holds if the radii coincide.
And if the radii coincide for every $s$, then the surfaces are the same and the metrics should be identical.
