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?
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.