An inverted cone of radius a and height h is filled with water. A sphere made of a material denser than water is placed in the cone. Find the radius of the sphere, r, that will displace the largest volume of water.
I pick up where B. Smith left off… He showed that the maximum displacement occurs for a sphere of radius R that is somewhere within the range of his Case 2 or Case 3. (Case 2 is where the sphere center is at or below the rim, but the sphere is not entirely submerged, and Case 3 is where the sphere is higher: its center is above the rim, but the sphere/cone contact circle is still at or below the rim.
He showed, for a sphere radius R, R is in these ranges:
Case 2, R = { ah / (a+sqrt(a^2+h^2), ah / sqrt(a^2+h^2) }
Case 3, R = { ah / sqrt(a^2+h^2), (a/h) sqrt(a^2+h^2) }
For every R, the sphere is immersed up to its spherical cap that sits above the rim (i.e. above the water). The volume displaced is the volume of the sphere up to the cap. If we picture the sphere in cylindrical coordinates centered at the origin, the underwater part extends in z from -R to z2. We have calculated z2 from R for Case 2 and Case 3 here (where we call it alpha).
z2 = h - (R/a) sqrt(a^2 + h^2)
In Case 2, z2 is positive, and in Case 3, z2 is negative.
We here calculate the volume displaced using z2 as a variable:
V = Int ( Int ( Int dV d_theta ) r d_r ) d_z
V = pi ((R^2 (z2 +R) - (1/3) (h - (r/a)c)^3 + R^3). where c = sqrt (h^2 + z^2)
In the calculation we show V has a maximum at:
R = - a c h / (2 a^2 - ac - c^2) with c = sqrt (a^2 + h^2)
For two cases I explored, a=3, h=6 and a=2, h=8, this is near the upper extent of Case 2, where the center of the sphere is just a bit below the rim.
This makes sense since almost half of the sphere is immersed and so it best approximates the shape of the cone. Were it located lower down, it would have to be quite diminished in radius in order to fit within the narrowing point of the cone.
These two examples are listed for various R showing the maximum radius R (from both the formula and empirically, here and here.
Edited on October 3, 2019, 1:00 am