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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 Solution (given my assumptions) | Comment 3 of 13 |

The sphere has radius R
The cylindar has radius r

First calculate the volume of the empty space of the cylindrical hole drilled through the sphere.

Use cylindrical shells.  Each shell has inner radius p, and outer radius p+dp.  To figure the height of each shell, picture a line from the center of the sphere to the extreme tip of the shell.  That line has length R, the radius of the shell is p, so the height of the shell is sqrt(R^2 - p^2).  But that's only the cylindar from the center of the sphere to the top, so multiply by 2 since the hole goes all the way through the sphere.
Volume of each cylindrical shell is: 
2*sqrt(R^2 - p^2)*(2 pi p dp)
p varies between 0 and r

Volume(cyl)= 4 pi integ(0 to r) {p*sqrt(R^2 - p^2)  dp}

substitute s=(R^2 - p^2)
            ds = -2p dp
Volume(cyl)= -2 pi integ(limits?) {sqrt(s) ds}
    where limits? are the limits redefined for s
    and s goes from R^2 to R^2 - r^2
Volume(cyl)= -(2 pi s^3/2 )/ (3/2) = - (4/3) pi s^(3/2)
Volume(cyl)= (4 pi / 3) * [R^3 - (R^2 - r^2)^(3/2)]

note that Volume of the sphere is (4 pi / 3) * R^3
so Volume(cyl)=Volume(sphere) - Volume(solid.that.remains)
this last is the volume we want, call it Volume(str)
Volume(str)=(4 pi / 3) *(R^2 - r^2)^(3/2)

But we know that R^2 - r^2 is half the height of the cylindar (not counting the rounded bits at each end):
cylindar height = 4 cm
R^2 - r^2 = 2 cm
(R^2 - r^2)^(3/2) = 8

Volume(str)=(32 pi / 3)
Which is independent of R, independent of the size of the sphere; and it is also independent of r, the diameter of the hole.  It does depend on the height of the cylindar that remains, defined by the edges of the hole.

  Posted by Larry on 2004-08-21 09:49:55
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