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Hole in a bead (Posted on 2004-08-21) Difficulty: 4 of 5
A round hole is drilled through the center of a spherical solid with a radius (r). The resulting cylindrical hole has height 4 cm.

a)What is the volume of the solid that remains?

b)What is unusual about the answer?

See The Solution Submitted by Pieater    
Rating: 3.0909 (11 votes)

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Solution Segmentation, not calculus | Comment 7 of 13 |

I will use the following variables.
r = radius of sphere
a = radius of cylindar
H = HALF of the height of the hole. So here, H = 2.
h = height of the spherical cap that had been removed off the ends of the cylinder. I hope that made sense.

Simple relationships:
h = r – H
r^2 = a^2 + H^2 so a^2 = r^2 – H^2

Sphere = 4/3*pi*r^3
Cylinder = pi*a^2*(2H) = 2*pi*a^2*H
Cap = 1/3*pi*h^2(3r–h)

Let’s find all of those equations just in terms of r and H, since that is all that we are really given.
Cylinder = 2*pi*a^2*H = 2*pi*H*(r^2 – H^2)

Cap = 1/3*pi*(r – H)^2*[3r – (r – H)]
       = 1/3*pi*(r^2 – 2rH + H^2)(2r + H)
       = 1/3*pi*(2r^3 + r^2*H – 4r^2*H – 2r*H^2 + 2r*H^2 + H^2)
       = 1/3*pi*(2r^3 –3r^2*H + H^3)

Alrighty, now the volume of the remaining solid is simply:
Solid volume = (V Sphere) – (V Cylinder) – 2*(V Spherical Cap)
V = 4/3*pi*r^3 – 2*pi*H(r^2 – H^2) – 2/3*pi*(2r^3 – 3r^2*H + H^3)
   = pi/3[4r^3 – 6H(r^2 – H^2) – 2(2r^3 –3r^2*H + H^3)]
   = pi/3[4r^3 – 6H*r^2 + 6H^3 – 4r^3 + 6r^2*H – 2H^3]
   = pi/3(4H^3) = 4/3*pi*H^3

So the answers are

  1. 4/3*pi*H^3 = 4/3*pi*(2cm)^3 = 32/3*pi cm^3= 33.51 cm^3

  2. I don’t know if it is unusual, but the volume of the solid doesn’t depend at all on the radius of the initial sphere. As the radius of the sphere gets larger, you are just taking out a much larger cap than before (and not just proportionally larger, I don’t think). Also, the volume of the solid left over is equivalent to a sphere with radius = H, which is interesting if not unusual.

  Posted by nikki on 2004-08-23 10:45:33
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