we can embed the portion $(r_1,\infty)$ just to see how the two-surface looks like. In this case, Eq. (\[30012014c\]) can be written in the form $$\frac{dz}{dR}=\frac{\sqrt{W-W_m^2}}{W_m}\quad \Rightarrow \quad  \frac{dz}{dr}=(W-W_m^2)^{1/2}W^{-(1+S)/2}, \label{24062014n}$$ where we have used the chain rule and $\pm |W_m|=W_m$ (the negative values of $W_m$ represent the region $z<0$).

Due to the cylindrical symmetry, we can parametrize the two-surface $t=constant$ and $\theta=\pi/2$ as $$\chi(r,\phi)=(R(r),\phi,z(R(r))). \label{26062014a}$$ The solution of Eq. (\[24062014n\]) will give us the explicit form of $\chi$.

For $S=3$, we manage to obtain the following exact solution: $$z=r_0\arctan \left(\frac{\sqrt{3r_0r-4r^2_0}}{r_0}\right)+\frac{2r-3r_0}{r-r_0}\sqrt{3r_0r-4r^2_0}-\left(\arctan\sqrt{2}+\sqrt{2} \right)r_0, \label{24062014p}$$ where the constant of integration has been chosen in such a way that the throat is at $z=0$. By using Eqs. (\[10082013f\]) and (\[24062014p\]) in Eq. (\[26062014a\]), one obtains the plot in Fig. \[24062014r\].

![In this figure we exhibit the plot of the two-surface (\[26062014a\]) with $r_0=1$ and $r$ varying from $4/3$ to $8$. As one can see, this surface has the shape of a typical wormhole. []{data-label="24062014r"}](f26062014a.jpg)

With the help of a computer, we have verified that the case $S=3/2$ yields a two-surface similar to the one in Fig. \[24062014r\]. This case is different from the previous one because $r_0$ is a physical singularity for $1<S<2$ (no 