All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Hole in a bead (Posted on 2004-08-21)
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?

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

Comments: ( Back to comment list | You must be logged in to post comments.)
 Segmentation, not calculus | Comment 7 of 13 |

I will use the following variables.
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

Volumes:
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

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

 Search: Search body:
Forums (1)