Our strategy for laying out the tubes will be to make them all drop down more or less vertically toward the (x,y)-plane. To get started, we will make all of the tubes vertical above height z=1. (See Figure 7.)
Figure 7: Cutting vertically down to height 1.
This is okay because all of the circles have radius 
,
so no hemisphere protrudes above z=1/2.
Below z=1,
we will be forced to bend the tubes.
In so doing,
we would like to make sure that the tubes stay more or less vertical,
in the sense that the angle
that the tubes make with the vertical stays bounded
above by some universal constant 
.
This way, the true cross-section of a tube will be more or less the same as
its horizontal cross-section,
that is, the area of its intersection
with the plane 
.
Similarly, the length (either hyperbolic or Euclidean)
of a section of tube will be
more or less the same as that of its projection
onto the z-axis.
Thus, instead of worrying about the growth of the cross-section as a function
of length along the tube,
we can worry about the growth of the horizontal cross-section
as a function of height above the (x,y)-plane.
In particular, if we can arrange things so that the tubes stay more or less
vertical,
and so that the horizontal cross-section of every tube never decreases,
and increases by a definite factor every time the vertical distance to the
(x,y)-plane drops by a factor of 2 (or 8, or whatever),
then the tubes will be growing more or less exponentially.
Of course near the tops of the hemispheres there is no way to keep the tubes more or less vertical. This is a real nuisance. To get around this difficulty, we will outfit each hemisphere with a duncecap, as shown in Figure 8.
Figure 8: Duncecap dimensions.
A hemisphere of radius a gets a cap whose apex is at height z=2a
and whose brim rests along the parallel at height z=a/2.
Call the region lying on or above the spruced-up hemispheres G.
If we can cut G up into tubes that grow more or less exponentially,
then the same goes for 
.
(See Figure 9.)
One way to see this is to consider that the obvious ``squash the hats''
map from G to taking (x,y,z) to (x,y,z-f(x,y))
is a quasi-isometry from the point of view of the hyperbolic metric.
So from now on we will concentrate on cutting up G,
rather than .
