Question 1018215
A cylindrical can has a radius of 3.4 cm.and when a sphere is placed in it, the water covers it.
To make this problem possible, we have to assume the sphere just fit in the cylinder. 
Three sides of sphere touched 3 sides of the cylinder
The top of the sphere was even with the top of the cylinder
Therefore the height of the cylinder and the diameter of the sphere was 6.8 cm
Twice the radius
 Now if the sphere is fit in it then find the:
:
A) total surface area of can. (including the bottom)
S.A. = {{{2*pi*3.4*6.8}}} + {{{pi*3.4^2}}} = 
S.A. = 145.267 + 36.317 
S.A. = 181.584 sq/cm
: 
B) depth of water in can before the sphere. 
Find the vol of the cylinder
V = {{{pi*3.4^2*6.4}}}
V = 246.954 cu/cm
Find the vol of the sphere
V = {{{4/3}}}{{{pi*3.4^3}}}
V = 164.636
Find the vol of the water when the sphere is placed in the cylinder
246.954 - 164.636 = 82.318 cu/cm of water
Find how high (h) the water will be when the cylinder is removed
{{{pi*3.4^2*h}}} = 82.318
36.317h = 82.318
h = {{{82.318/36.317}}}
h = 2.267 cm is the height of the water in the cylinder when sphere is removed