Consider a hemisphere 
,
associated to the circle C and the disk D.
Let the radius of C be a.
We assume that is relevant,
and not a baby,
that is,
that a > 2h.
At height h,
the tubes trapped over cover an annulus A(h) in the plane
of outer radius a
and inner radius
.
(See Figure 11.)
Figure 11: No room at infinity reprise.
The Euclidean area of A(h) is
This annulus is the image under projection into the plane
of the annulus 
consisting the intersection of
the plane z=h with the locus of points above
and inside the cylinder over C.
Earlier we determined that the hyperbolic area of
was on the order of 1.
Now we see that it is exactly pi.
In order to insure that the tubes over
have enough room to grow,
we will need to annex around the base of the hemisphere
a region B
having Euclidean area on the order of that of A (h),
that is,
we must have
for some universal constant K.
To see that this is the right condition,
let 
be the portion of the plane z=h
that lies over B.
The hyperbolic area of 
is
while the hyperbolic area of 
is
Our assumption on the area of B guarantees that the ratio of
the latter quantity to the former will always be 
.
Thus in going from height h down to height h/2 the hyperbolic
area available to the tubes increases by a definite factor,
which is the kind of condition we need.