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## The Famous Hole in a Sphere 8

This incredible problem was first made famous by Martin Gardner with his ‘Mathematical Games’ column published in Scientific American throughout the 60s and 70s.

A cylindrical hole, 6 inches long, has been drilled straight through the centre of a solid sphere. What is the volume remaining in the sphere?

No additional data is needed to solve this. (really!)

36pi inch^3

Without using calculus, one of the key equations to solving this problem is the volume of a spherical “cap”. That is the volume above a plane that has bisected a sphere. It is (1/6)πh(3r2 + h2) (where h is the altitude of the cap, and r is its radius). (See the mathworld page for more info)

The volume remaining in the sphere will be the volume of the sphere – (the volume of the “hole” + 2*(volume of the cap)). Refer to the drawing below for clarification.

The volume of the sphere is simply 4/3πR3. The volume of the hole is 6π(R2 – 9). The volume of both of the caps can be obtained by plugging in h = R – 3, and r = sqrt(R2 – 9)) .

After a long, boring simplification, we find that all the terms cancel to leave 36pi!

Try it for yourself.

leilei3915

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leilei3915

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Sean Moran

You could just do: The diameter d of the sphere is 6 in (because the cylinder is 6 in long & it goes through the middle of the sphere, i.e. the part that has the "true" diameter). Therefore, the radius r is 3 in (r = d / 2 = 6 / 2 = 3). Then, using the equation V = 4/3pi(r^3) we get 36pi. Also note the cylindrical hole doesn't take any volume away, so the volume before the hole is taken out is the same as the volume after.

Ravi

I don't think that's true. The question is asking what the remaining volume of the solid sphere is after the cylinder is removed.

Also, check out this picture. Notice that the height h of the cylinder does not equal the diameter of the sphere.

Sean Moran

Volume is the measure of how much space there is in an area, not how much stuff is in there. The unit of volume is length cubed, which implies this as well. Therefore, a solid sphere has the same amount of volume as the same area of vacuum or air. As for your second point, the poster said the cylinder was drilled straight through the middle, and by definition a line straight through the middle of a circle or sphere is the diameter. Also, using my method, I got the same answer as the poster got.

Sean Moran

Actually, I checked out the picture and you might be right, but the poster only said the length of the cylinder, not radius or circumference or anything, so the answer would not be known unless the radius were so small as to not make the difference matter much. Also, using my method you still get the same amount, so apparently the source didn't think about it. I still keep my comment about volume, though.

Tarun245

So is the diameter 6 inches?

Nigel Parsons

No, (or not necessarily). The puzzle doesn't state that. To drill a hole 6in long through the centre of the sphere, the sphere must have a minimum diameter of 6 inches. In which case, the hole would be of negligible diameter (and the remainder would effectively still be a 6 inch sphere) as otherwise the hole would be less that 6 inches long. If you start with a sphere of 7 inches then for the hole to go through it, and be only 6 inches long the hole would have to be wide enough so that the 'caps' taken off each end have a maximum thickness (at the centre) of 1/2 inch each. This will mean that the larger the sphere, the wider must be the 6 inch hole in order for it to be only 6 inches in length. As such the result can be calculated without knowing the size of the sphere, strange as that may sound.

viswa

if 1 * 2 < 3:

``````    print "hello, world!"
``````

JWE

No the diameter is not 6 inches To have a discrete answer the problem must therefore be to evaluate the volume of the sphere else the answer would determined by the diameter of the cylindrical hole and ought to be asking for the remaining solid volume.

Trey Klein

As the diameter of the hole approaches zero, the answer must approach the diameter of the sphere: thus the answer is 4/3 * pi * radius * radius * radius... (the volume of a sphere) which equals 36 pi in this case. How is this this not the easiest and most logical answer to the puzzle? You don't need to know (or even understand) the concept of the spherical cap or the volume of the hole!

Trey Klein

(sigh) I mean the answer must approach the VOLUME of the sphere! It works!

Trey Klein

(sigh) I mean the answer must approach the VOLUME of the sphere! It works!

Trey Klein

(sigh)

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Submitted by
Ravi
almost 9 years ago
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Difficulty 6.7 ?

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