Manifold with minimum surface distance between two points The book "The World is Flat" uses flatness as a metaphor for a global economy. In fact, a spherical world would seem to be better than a flat world in terms of reducing the distances between two random points on the surface of the world. The shorter the distance between any two points, the easier it is for information and objects to travel between different places.    While it may be obvious that a spherical world is better than a flat world, it's far from obvious that a spherical world is optimal in this regard, which brings me to the question: what should the book have been titled?  More precisely:
Question: Define a world to be a 2-manifold with some fixed surface area S and a metric d that calculates distance on the surface of the manifold. What shaped world minimizes average distance between any two randomly selected points in the world? 
Does the answer depend on whether the world can be embedded in $\mathbb{R}^3$?  
Does it depend on the specific metric used?
Does it depend on how we define "average distance" or "randomly selected?"
 A: The average distance on a sphere of radius $r$ is $\frac{\pi}{2}r$ (which we can get without any integrations because there's as much area at a distance $\frac{\pi}{2}+\theta$ from a given point as there is at $\frac{\pi}{2}-\theta$); so this is
$$\bar{d}=\frac{\pi}{2}\sqrt{\frac{A}{4\pi}}=\frac{\sqrt{\pi}}{4}\sqrt{A}\approx 0.443 \sqrt{A}\;.$$
The average distance on the flat manifold $(\mathbb R / a\mathbb Z)^2$ is $\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)a$ (the average Euclidean distance between the point $(a/2,a/2)$ and a random point in $[0,a]^2$), and the area of that manifold is $a^2$, so that makes
$$\bar{d}=\frac{1}{6}(\sqrt{2}+\sinh^{-1} 1)\sqrt{A}\approx 0.383 \sqrt{A}\;.$$
So it's not true that a flat manifold has higher average distance than a sphere; it's the other way around. I suspect the comparison you mean is between a sphere and a flat disk; but a disk is a manifold with boundary, not a manifold, and the higher average distance is due to the boundary, not to the flatness.
[Edit:] Here's the calculation of the average distance in the flat case, as requested. [This is a corrected version.]
We'll need an integral of the form $\sqrt{x^2+c^2}$ several times, so I'll do that in general form first, using the substitution $x=c\sinh u$:
$$
\begin{eqnarray}
\int\sqrt{x^2+c^2}\mathrm dx
&=&
\int c\sqrt{1+\sinh^2 u}\;c\cosh u\mathrm du
\\
&=&
c^2\int\cosh^2u\mathrm du
\\
&=&
\frac{c^2}{2}(u+\sinh u\cosh u)
\\
&=&
\frac{c^2}{2}\left(\sinh^{-1}\frac{x}{c}+\frac{x}{c}\sqrt{1+\left(\frac{x}{c}\right)^2}\right)
\\
&=&
\frac{1}{2}\left(c^2\sinh^{-1}\frac{x}{c}+x\sqrt{x^2+c^2}\right)\;.
\end{eqnarray}
$$
Then the integral over the distance in the primitive cell of a square lattice with $a=2$ is
$$
\begin{eqnarray}
\int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy
&=&
4\int_0^1\int_0^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy
\\
&=&
4\int_0^1\left[
\frac{1}{2}\left(y^2\sinh^{-1}\frac{x}{y}+x\sqrt{x^2+y^2}\right)
\right]_0^1 \mathrm dy
\\
&=&
2\int_0^1
\left(y^2\sinh^{-1}\frac{1}{y}+\sqrt{y^2+1}\right)
\mathrm dy\;.
\end{eqnarray}
$$
We can deal with the first term by integrating by parts twice:
$$
\begin{eqnarray}
\int y^2\sinh^{-1}\frac{1}{y}\mathrm dy
&=&
\frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{1}{\sqrt{1+(1/y)^2}}\mathrm dy\right)
\\
&=&
\frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+\int y\frac{y}{\sqrt{y^2+1}}\mathrm dy\right)
\\
&=&
\frac{1}{3}\left(y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}-\int \sqrt{y^2+1}\mathrm dy\right)\;.
\end{eqnarray}
$$
Putting everything together, we get
$$
\begin{eqnarray}
\int_{-1}^1\int_{-1}^1\sqrt{x^2+y^2}\mathrm dx \mathrm dy
&=&
2\left\{\frac{1}{3}\left[
y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}
\right]_0^1
+\left(1-\frac{1}{3}\right)
\int_0^1\sqrt{y^2+1}\mathrm dy\right\}
\\
&=&
\frac{2}{3}\left[
y^3\sinh^{-1}\frac{1}{y}+y\sqrt{y^2+1}
+\sinh^{-1}y+y\sqrt{y^2+1}\right]_0^1
\\
&=&
\frac{4}{3}\left(\sinh^{-1}1+\sqrt{2}\right)\;.
\end{eqnarray}
$$
This has to be divided by the area $2^2$ to get the average distance for $a=2$, and then by $2$ to get the average distance for $a=1$, since the average distance scales linearly with $a$; that leads to the above result.
For the average distance on the quotient of the plane with respect to a hexagonal lattice, we can use symmetry to restrict the calculation to half of one of the $6$ equilateral triangles. The integrals are rather complicated to work out by hand, but can be done analytically; WolframAlpha gives
$$
\int_0^{\sqrt{3}/2} \int_0^{1-y/\sqrt{3}} \sqrt{x^2+y^2}\mathrm dx \mathrm dy
=\frac{4+\log 27}{32\sqrt{3}}
$$
for side length $a=1$. The area of one of these half-triangles is $\frac{\sqrt{3}}{8} a^2$, so the total area of the manifold is $\frac{3\sqrt{3}}{2}a^2$, and the average distance comes out as
$$\bar{d}=\left(\frac{4+\log 27}{32\sqrt{3}}/\frac{\sqrt{3}}{8}\right)a=\frac{4+\log 27}{12}\sqrt{\frac{A}{3\sqrt{3}/2}}\approx 0.370 \sqrt{A}\;,$$
so yasmar's manifold indeed slightly improves on the one using a square lattice.
