[net.math] Correct 6-inch hole solution

drabik (08/06/82)

The solution given in Article 171 by pur-ee!jarvis is incorrect.
The length of a cylindrical hole through a sphere decreases as the
diameter of the hole increases.  If the length of the hole is 6 and
the (arbitrary) radius of the sphere is r, then the radius of the
cylindrical hole is

	sqrt (r**2 - 9)

Let the remaining volume be sliced into rings by planes
perpendicular to the axis of the cylinder.  A section formed by the
intersection of the volume and the plane a distance x from the
center of the sphere has area

	pi * (r**2 - x**2) - pi * (r**2 - 3**2)

or

	pi * (9 - x**2)

Integrating, we get


             3
             /
	V =  ! pi * (9 - x**2) dx  =  36 * pi
             /
           x = -3

Note that the integrand is independent of r.  We need not evaluate
the integral.  The volume can be determined by letting r = 3, and
finding the volume of a sphere of radius three minus the volume of a
cylinder of radius zero, i.e.

	V = (4/3) * pi * 3**3 = 36 * pi

I am at a loss, however, for a physical interpretation of this
situation.

Tim Drabik
Bell Labs New Switching Services Laboratory
Naperville, Illinois, USA
RHIOT '81

mark@sri-unix (08/19/82)

#R:zehntel:-36800:beta:5400001:000:140
beta!mark    Aug 11 10:23:00 1982

I first ran across this problem in a beginning calculus final.
It's a favorite of mine, too.

Mark Wittenberg
...!decvax!sytek!zehntel!mark