Find the curvature tensor and sectional curvature associated with the first fundamental form$I=du^2+f^2(u)dv^2$ Consider the surface of revolution $\sigma$ in he Euclidean space $\mathbb{R^3}$ given by $$\sigma(u,v)=(f(u)cosv,f(u)sinv, g(u))$$ with $f>0$ where the profile curve has unit speed. 
The first fundamental form associated with the surface patch $\sigma$ is given by $$I=du^2+f^2(u)dv^2$$
I am trying to calculate the curvature tensor and the sectional curvature associated with this first fundamental form. 
This is the proof.
For the Christoffel symbols of the first kind, they are all non zero except $$\Gamma_{221}=-f(u)f'(u)$$ and $$\Gamma_{212}=\Gamma_{122}=-\Gamma_{221}$$
For the Christoffel symbols of the second kind we have $$\Gamma^2_{21}=\Gamma^2_{12}=\frac{f'(u)}{f(u)}$$ and $$\Gamma^1_{22}=-f(u)f'(u)$$

From this how do we get that the Reimann curvature tensor is
  $-f(u)f''(u)$?

I am told that using the symmetries of the Reimann curvature tensor we obtain $$K(x)=\frac{-f''(u)}{f(u)}$$

How do we find the curvature explicitly?

 A: Just FYI, there is a formula for computing Gaussian curvature of a surface orthogonally coordinated. Suppose the first fundamental form is 
$$
ds^2 = A^2du^2 + B^2dv^2
$$
where $A, B$ are functions of $u, v$. Then the Gaussian curvature is 
$$
K = - \frac {1}{AB}\left(\left(\frac{A_v}{B}\right)_v + \left(\frac{B_u}{A}\right)_u\right).
$$
The subscript here just means taking partial derivatives. If the coordinate is isothermal moreover, the formula reduces to the familiar one
$$
K = -\frac{1}{2g}\Delta \log g.
$$
These formulae can be proved by assuming a local imbedding into $\mathbb R^3$ (which you can find in GTM 93), but I will now compute $K$ by contracting the Riemann tensor.
We only need to compute $R^v_{uvu} = R_{uu}$, which is
$$
-\left(\frac{AA_v}{B^2}\right)_v-\left(\frac{B_u}{B}\right)_u + \frac{A_uB_u}{AB} - \frac{AA_vB_v}{B^3} + \frac{A_v^2}{B^2} - \frac{B_u^2}{B^2} = -\frac{A^2}{AB}\left(\left(\frac{A_v}{B}\right)_v + \left(\frac{B_u}{A}\right)_u\right)
$$
Therefore
$$
R = g^{uu}R_{uu} + (u \leftrightarrow v) = - \frac {2}{AB}\left(\left(\frac{A_v}{B}\right)_v + \left(\frac{B_u}{A}\right)_u\right) = 2K.
$$
