SOLUTION: Two metal balls are placed in a cylinder containing water. The lower ball has a radius of 7 cm, and the upper ball has a radius of 3 cm, and the diameter of the cylinder is 18 cm.

There are two basic stable configurations for the spherical balls in this cylinder.

 
  

  

One configuration is shown in the Figure on the right.


The larger ball lies at the bottom of the cylinder and touches the cylinder's vertical lateral surface.   


The smaller ball is above the bottom. It touches the larger ball and also touches vertical lateral 
surface of the cylinder.


The Figure shows a vertical section of the cylinder and the two balls by that existing and unique 
vertical plane which contains the cylinder's vertical axis and the centers of the two balls.


In the Figure, you see a right angled triangle shown in red.


The hypotenuse of this triangle is the segment AB connecting the centers of balls.

This segment goes through the tangent point, and its length is  7 + 3 = 10 cm.


You can easily find the horizontal leg of this triangle, x.

From the equation  7 + x + 3 = 18  you have  x = 18 - 7 - 3 = 8.

 
  




  
  
 
Then the vertical leg of this triangle is   =  = 6 cm.


Thus the height of the highest point of the small ball over the cylinder bottom is  7 + 6 + 3 = 16 centimeters.


It defines the minimal required upper level of 16 cm of the water in the cylinder after submerging the balls.


With it, the volume equation is


 = W +   + ,   or


W =  -  =  =  = 


where W is the minimal required volume of the water before submerging the balls.


Now, how much of water has to be poured out so that there is just enough water to cover both balls? - It is the difference 


     cm^3  MINUS   cm^3 =  cm^3 =  cm^3.     ANSWER