Total Curvature of 4 pi What does it mean for a surface to have a total curvature of $4\pi $? 

I have seen that both the catenoid and Enneper surface are the only minimal surfaces that have this total curvature, but I don't really understand what significance this has?

Could anyone please explain this?
For calculate the finite total curvature of a complete minimal surface $\phi:M\longrightarrow\mathbb{R}^3$, by start from the topology of $\bar{M}$ and the geometry of ends.

An $\mathbf{end}$ of a completle minimal surface of finite total curvature is the image $c_i=\phi(D_i-p_i)$ of a sufficiently small punctured disk $D_i-p_i$ on $\bar{M}$ with centered at the point $p_i$.

$\mathbf{Theorem}$. If $g$ is the genus of $\bar{M}$, $M\cong{\bar{M}-\{p_1,...,p_r\}}$
and $d_1,...,d_r$ are the multiplicities of the ends $c_1,...,c_r$, and $\tau$ the total curvature

$$\tau(M)=2\pi(2-2g-r-\sum_{j=1}^rd_j)=2\pi(\chi(\bar{M})-r-\sum_{j=1}^rd_j).$$

--M=Catenoid has two ends. $r=2, d_1=d_2=1$, and $\bar{M}=\mathbb{S}^2$. Consequently

$$\tau(M)=2\pi(2-2-2)=-4\pi.$$

--The M=Enneper has one end of multiplicitiy $d_1$, and $\bar{M}=\mathbb{S}^2$, $\tau(M)=-4\pi$ Consequently $d_1=3$.
Well first of all these have total curvature $-4\pi$, not $4\pi$.

The Gauss-Bonnet theorem tells us that the total curvature of a surface is equal to $2\pi$ times the Euler characteristic of the surface. Moreover we know that the Euler characteristic of a connected surface is always an integer that is at most 2.

So in general we might want to ask: What are all connected minimal surfaces without boundary in $\mathbb{R^3}$ of with total curvature $2\pi n$? We see this is the same as asking that the Euler characteristic be n.

Well, if $n=1,2$ the answer is none, since a minimal surface has nonpositive Gaussian curvature everywhere. Now we can go down from there trying to understand the most topologically simple minimal surfaces in $\mathbb{R^3}$.

For $n=0$ it's not hard to figure out that a plane is the only connected minimal surface without boundary in $\mathbb{R^3}$ that has total curvature 0.

What you are saying is that these are the only two minimal surfaces in $\mathbb{R^3}$ with Euler characteristic $-2$. So we see this case still has a relatively nice answer.

But as we decrease $n$ even further the story becomes increasingly complicated, and a general answer is unknown.