Question 1189457
<pre>
Unlike the other tutor, I will assume the ball is heavy 
enough to sink all the way down and touch the sides of 
the cone.  No offense, Greenestamps! lol

Here is the way it looks:

{{{drawing(800,4800/13,-13,13,-9,3, circle(0,sqrt(13)-8,3),
triangle(-12,0,12,0,0,-8), line(0,sqrt(13)-8,1.664100589,-6.890599608),
line(0,sqrt(13)-8,-1.664100589,-6.890599608),
green(line(0,0,0,-8)), locate(0,-8,C),locate(-1.9,-6.9,A),locate(1.7,-6.9,B),
locate(0.1,-3.86,O),locate(.1,-6.4,P),
green(line(-1.664100589,-6.890599608,1.664100589,-6.890599608)),
locate(-12,.6,D), locate(12,.6,E), red(locate(0,2,24),line(-12,2,12,2),
line(13,0,13,-8)),locate(1,-5.3,3),locate(-1.2,-5.3,3),locate(6,0,12),locate(-6,0,12),

arc(0,sqrt(13)-8,6,-1.5,180,360),green(arc(0,sqrt(13)-8,6,-1.5,0,180)),
arc(0,sqrt(13)-8,1.5,-6,90,270),

locate(-.3,-7.3,Q),

arc(0,-6.890599608,2*1.664100589,-.6,180,360),
green(arc(0,-6.890599608,2*1.664100589,-.6,0,180)),

locate(0,.5,F), red(locate(12.55,-4,8)))}}}

It's easy to see that all 6 right triangles DFC, EFC, OPA, OPB, 
APC, and BPC are similar. Here's a mid-cross-section 

{{{drawing(800,4800/13,-13,13,-9,3, circle(0,sqrt(13)-8,3),
triangle(-12,0,12,0,0,-8), line(0,sqrt(13)-8,1.664100589,-6.890599608),
line(0,sqrt(13)-8,-1.664100589,-6.890599608),
green(line(0,0,0,-8)), locate(0,-8,C),locate(-1.9,-6.9,A),locate(1.7,-6.9,B),
locate(0.1,-3.86,O),locate(.1,-6.4,P),
green(line(-1.664100589,-6.890599608,1.664100589,-6.890599608)),
locate(-12,.6,D), locate(12,.6,E), red(locate(0,2,24),line(-12,2,12,2),
line(13,0,13,-8)),locate(1,-5.3,3),locate(-1.2,-5.3,3),locate(6,0,12),locate(-6,0,12),
locate(0,.5,F), red(locate(12.55,-4,8)),  locate(-.3,-7.3,Q)

 

 )}}}

By using the Pythagorean theorem to calculate the hypotenuse CD = 4√(13),
we see that this ratio holds for all 6 right triangles 

shorter leg : longer leg : hypotenuse = 8 : 12 : {{{4sqrt(13)}}}

Dividing through by 4

shorter leg : longer leg : hypotenuse = 2 : 3 : {{{sqrt(13)}}}

Below we will draw the kite ACBO with all the lengths which
we calculate by ratio and proportion of similar triangles
(You don't need to calculate them all).

{{{drawing(1120/3,400,-7,7,-5,10,  locate(-.3,-1.1,Q),

green(arc(0,9,2sqrt(117),-2sqrt(117),237,304)),
locate(-6.5,.5,A), locate(6.29,.5,B),locate(0,9.6,O),locate(.1,.6,P),
locate(3.4,4.7,3), locate(-3.63,4.7,3),locate(-3,1.6,6/sqrt(13)),
locate(2.8,1.6,6/sqrt(13)),locate(.1,4.5,9/sqrt(13)),locate(0,-4,C),
locate(.1,-1.2,4/sqrt(13)),locate(-3.4,-1.9,2),locate(3.3,-1.9,2),
triangle(-6,0,6,0,0,9), triangle(-6,0,6,0,0,-4), triangle(0,-4,0,9,0,0))}}}

So the radius of the cone at the bottom (that the sphere "sits in") is 

{{{AP}}}{{{""=""}}}{{{6/sqrt(13)}}}

and the height of that cone is

{{{CP}}}{{{""=""}}}{{{4/sqrt(13)}}}

We compute its volume by the formula:

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

{{{V}}}{{{""=""}}}{{{expr(1/3)*pi*(6/sqrt(13))^2(4/sqrt(13))}}}{{{""=""}}}{{{expr(1/3)*pi*(36/13)(4/sqrt(13))}}}{{{""=""}}}{{{(48pi)/(13sqrt(13))}}}

However, we see that the volume we want is not the entire cone, because the cone
is "dished out" by the sphere sitting on it.  The "dished out" part is a cap of
the sphere.  So we must calculate the volume of the cap of the sphere which
dishes out the cone, so we can subtract it from the above result.  The formula
for the volume of the cap of a sphere is

{{{V}}}{{{""=""}}}{{{expr(1/3)pi*h^2(3r-h)}}} 

where r = radius of the sphere = 3, and 
h = height of the cap.

We calculate the height of the cap, PQ:

PQ = OQ - OP.  OQ is a radius of the sphere, so OQ = 3. So

{{{PQ}}}{{{""=""}}}{{{3-9/sqrt(13)}}}  

So the volume of the cap is

{{{V}}}{{{""=""}}}{{{expr(1/3)pi*(3-9/sqrt(13))^2(3(6/sqrt(13))-(3-9/sqrt(13)))}}}

That simplifies to {{{expr(36pi/169)(69sqrt(13)-247)}}}

So we subtract that from  {{{(48pi)/(13sqrt(13))}}} and get the
volume of water below the sphere as

{{{(48pi)/(13sqrt(13))}}}{{{""-""}}}{{{expr(36pi/169)(69sqrt(13)-247)}}}

which simplifies to:

{{{expr(12/169)pi*(793 - 207sqrt(13))}}} cubic centimeters.

Edwin</pre>