Marilyn Comes up Short on the Hole through the Sphere

Ask Marilyn ® by Marilyn vos Savant is a column in Parade Magazine, published by PARADE, 711 Third Avenue, New York, NY 10017, USA. According to Parade, Marilyn vos Savant is listed in the "Guinness Book of World Records Hall of Fame" for "Highest IQ."

In her Parade Magazine column ofOctober 20, 1996, Marilyn reported that if you bore a perfect six inch long cylindrical hole through the center of a solid sphere, you always wind up with 113.09724 cubic inches of sphere remaining, regardless of the diameter of the original sphere.

Three Minor Problems

There are three problems with Marilyn's answer:

• Marilyn is incorrect in referring to a "6-inch sphere." Since boring a hole through the sphere cuts off the rounded end at each end of the hole, and since the hole is 6 inches long, the diameter of the sphere must be larger than 6 inches. In fact, Marilyn knew that the size of the sphere varies; she said so in her previous sentence.

• Marilyn's answer of 113.09724 is slightly inaccurate. The correct answer is 113.09734. If we assume that Marilyn is correct that the size of the sphere doesn't matter, let's calculate the volume using an infinitesimally thin (or pinhole size) hole in a six inch sphere. The volume of the sphere is 4/3 x pi x r^3 (four-thirds pi r cubed). The radius (r) is 3 inches (half the diameter), and Pi is approximately 3.14159265358979. The problem is that Marilyn's answer specifies 8 significant digits, but she used a value of 3.14159 (only 6 significant digits) for pi when calculating her answer. The result is round off error in the seventh significant digit. Marilyn should have either specified fewer significant digits in the result, or used a more precise value for pi in her calculations.

• The third problem is that Marilyn omitted all the details of her solution, leaving many readers wondering how she could possibly be correct. Since many readers would be lost in the calculus, let's begin with an intuitive explanation.

  It is easy to visualize that an infinitesimally thin (or pinhole size) sized hole in a six inch sphere removes very little volume, leaving most of the sphere intact. Now, imagine what it takes to make a six inch hole in a 12 inch diameter sphere. First, grind three inches off the top and the bottom of the sphere, so that it is only six inches thick, and now drill a hole through the remaining portion as wide as the flat surfaces on the top and bottom. Clearly, this process has removed a substantial percentage of the original volume. Thus, although we began with a larger sphere, it is possible to visualize intuitively how the remaining volume may be no larger than that of the six inch sphere.

Professor Edward Gieszelmann's Solution

As I said, this is a first-semester calculus problem. The volume is a volume of revolution. It can be found by using the washer method to set up the integral for the volume. First, consider a slice through the sphere that is through the axis of the hole. Now you can draw a right triangle that has as its legs 1) half the axis of the hole, which is 3", starting at the center of the sphere and running to the end of the hole, and 2) the radius of the hole, call it r, at the end of the hole. The hypotenuse of this triangle is the radius of the sphere, R. Using the pythagorean theorem, R^2 = r^2 + 3^2, where ^ means exponentiation. On this slice, the equation of the circle that is revolved to form the sphere is x^2 + y^2 = R^2. Now, using the method of washers, the volume is

2*integral (from 0 to 3) [pi*(R^2 - x^2) - pi*r^2] dx

or

2*pi*integral (from 0 to 3) (R^2 - r^2 - x^2) dx

But, from the pythagorean theorem above R^2 - r^2 = 3^2, so that the integral becomes

2*pi*integral (from 0 to 3) (3^2 - x^2) dx

or 2*pi*[9x - (x^3)/3] evaluated between 0 and 3.

This gives 2*pi*[9*3 - 27/3] or 2*pi*[27 - 9] = 36*pi

Thus the volume of the remainder of the sphere is constant; i.e., it does not depend upon the radius of the sphere.

The physical modelling of this is somewhat difficult. One can observe that the remaining volume gets thinner and thinner as the radius of the sphere increases. The greatest thickness of the volume is R - r, where R and r are related by the pythagorean theorem above. For example, if r = 1000", then R^2 = 1000^2 + 3^2 = 1000009, and R = 1000.0045 approximately. This makes the remaining volume .0045" thick at its thickest point. Essentially all that is left of the sphere is a 6" wide ribbon. If I were to make the ribbon into one of uniform thickness, I'd get one about .003" thick, since the rounding would make it thicker than half of .0045". If I were to take such a ribbon and wind it around a 1000" radius cylinder, the volume of the uniform thickness ribbon would be the circumference of the cylinder times the width of the ribbon times its thickness, or a volume of 2*pi*1000*6*.003 or 36*pi cubic inches. I think if you work out a few examples, they'll all come close, and you'll see that the calculus is correct.

An Alternative Solution

The easiest(!) way to determine the volume is as a triple integral in cylindrical coordinates. The volume of a region is given by (I'll use S's for integral signs and @ for "theta"):

    SSS r dr d@ dz

where setting the bounds for each integral is the tricky part.

@ varies from 0 to 2Pi, and nothing depends on it, so the integral reduces to:

    2Pi SS r dr dz.

Now, z varies from -sqrt(R^2 - r^2) to +sqrt(R^2 - r^2), where R is the radius of the sphere, so, integrating over z, and simplifying, we now have

    4Pi S r * sqrt(R^2 - r^2) dr

where r varies from sqrt(R^2 - L^2) to R (for convenience, L is half the length of the hole, or 3" in the particular case cited).

Transforming coordinates v = R^2 - r^2, the integral becomes:

    -2Pi S sqrt(v) dv

where v varies from L^2 to 0.

Solving, we get:

    -4/3Pi v^3/2

evaluated from L^2 to 0, or:

    4/3Pi L^3,

which is, of course, the volume of a sphere of radius L.

Note that for L = 3, this is ~113 square inches, which was part of the original claim.