In the case of an axially symmetric system the nuclear surface is given by

$$\rho_s^2\,(z) = R_0^2 \, \sum_{\ell=0}^\Lambda \alpha_{\ell} \, 
                             P_{\ell}\,\left (\frac{z-{\bar z}}{z_0} \right ) 
        = R_0^2 \, \sum_{\ell=0}^\Lambda \alpha_{\ell} \, P_{\ell}\,(u) \; ,
                       \hspace{ 0.3cm}  z_0 = \frac{2 R_0}{3 \alpha_0} \; , 
                   \hspace{0.3cm} u = \frac{z-{\bar z}}{z_0}
\label{eqon1}$$

with $2 z_0$ the elongation of the shape in z direction, ${\bar z}$ its geometrical center and $R_0$ the radius of the corresponding spherical nucleus. The deformation parameters $\alpha_\ell$ define the shape.

This parametrization is strongly related to the well known Funny Hills $\{c,h, \alpha\}$ parametrization [@BDJ72] recalled below :

$$\rho_s^2\,(z) = c^2 \, R_0^2 \left\{ \begin{array}{ll}
   (1 - u^2) \, (A + \alpha \, u + B \, u^2) \; , \;            & B \geq 0  \\ 
   (1 - u^2) \, (A + \alpha \, u) \; {\rm exp}(B \, c^3 \, u^2) \; ,\; & B < 0 
\end{array}\right.
\label{eqon2}$$

with $z_0 = c \, R_0$ and where $A$ and $B$ are related to $c$ and $h$ through $$A = \frac{1}{c^3} - \frac{1}{5} B \; , \;\;\;\;\;\;\;\; 
     B = 2 \, h + \frac{1}{2} (c-1) \;\; .$$

We have tested the convergence of these parametrizations for the description 