Question 1205736
<pre>

First we look at the mid cross section.
Let the radius of the circle (which is also the radius of the sphere),
be "a".

{{{drawing(400,400,-3.5,3.5,-1,6,green(line(0,3.7,1.3,3.3)),
line(-2,5.03,0,0), line(2,5.03,0,0),line(-2,5.03,2,5.03),
blue(line(0,0,0,5.03)),locate(.55,3.55,a),
locate(-.2,3.9,O),locate(1.4,3.4,P),locate(-.1,0,Q),
locate(-.1,5.35,S),locate(1.9,5.35,T),locate(-.2,4.6,a),

circle(0,3.7,1.38) )}}}

Then the ratio of the volume of the sphere to the volume of the cone,
which is the answer you're looking for, will be

{{{ratio}}}{{{""=""}}}{{{(expr(4/3)pi*a^3)/(expr(1/3)pi*r^2*h)}}}{{{""=""}}}{{{4a^3/(r^2*h)}}}

The hard part is finding the radius of the sphere "a", in terms
of "r" and "h".

By similar right triangles OPQ and TSQ,

{{{OP/OQ= TS/TQ}}}

SQ = h
SO = a
OQ = SQ-SO = h-a
OP = a
TS = r
{{{TQ=sqrt(TS^2+SQ^2)}}}{{{""=""}}}{{{sqrt(r^2+h^2)}}}

Substituting,

{{{a^""/(h-a^"")}}}{{{""=""}}}{{{r^""/sqrt(r^2+h^2)}}} 

Cross-multiplying:

{{{a*sqrt(r^2+h^2)}}}{{{""=""}}}{{{r(h-a)}}}  

{{{a*sqrt(r^2+h^2)}}}{{{""=""}}}{{{rh-ra}}} 

{{{a*sqrt(r^2+h^2)+ra}}}{{{""=""}}}{{{rh}}}

{{{a(sqrt(r^2+h^2)+r)}}}{{{""=""}}}{{{rh}}}

{{{a}}}{{{""=""}}}{{{(rh)/(sqrt(r^2+h^2)+r)}}}
 
Now we substitute that horrible thing in: 

{{{ratio}}}{{{""=""}}}{{{4a^3/(r^2*h)}}}

{{{ratio}}}{{{""=""}}}{{{4((rh)/(sqrt(r^2+h^2)+r))^3/(r^2*h)}}}{{{""=""}}}{{{expr(4/(r^2*h)) 
((rh)/(sqrt(r^2+h^2)+r))^3}}}{{{""=""}}}{{{expr(4/(r^2*h)) (r^3h^3)/(sqrt(r^2+h^2)+r)^3}}}{{{""=""}}}{{{(4r*h^2)/(sqrt(r^2+h^2)+r)^3}}}

Edwin</pre>