z
We can find the volume of a sphere using the
following integral:
R
V    R 2  z 2 dz
h
R
R
R
 2
z3 
4
V    R z    R 3
3  R 3

-R
Now, to find the volume of the submerged part ( 4/5V ) in terms of the
integral then z will be from –R to h. Where h is the unknown height:
h
4
V    R 2  z 2 dz
R
5
h

4 4 3
z3 
 R    R 2 z  
5 3
3  R

16 3  2
h3  
R3 
3

R   R h      R 
15
3 
3 

16 3
h3 2 3
R  R2h 
 R
15
3 3
3
h
h
   3   1.2  0
R
R
Solving this equation using a calculator we get:
h
h
h
 1.479 OR
 1.905 OR
 0.426
R
R
R
The first solution is unacceptable because if h > R this means that the
whole sphere is submerged and this is not the case. The second solution is
also unacceptable because if h < 0 this means that less than half of the
sphere is submerged (exactly 1/5 of its volume) which is not the case.
h
  0.426
R