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Topic: Hole Through Sphere (Read 2173 times) |
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ThudnBlunder
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Hole Through Sphere
« on: Apr 13th, 2003, 5:14pm » |
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A hole 6" long is drilled through the centre of a solid sphere (whose diameter is, of course, at least 6").
What volume of the sphere's material remains?
(I searched for this puzzle but couldn't find it.)
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| « Last Edit: May 29th, 2004, 7:27pm by ThudnBlunder » |
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aero_guy
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Re: Hole Through Sphere
« Reply #1 on: Apr 13th, 2003, 5:19pm » |
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Wouldn't that depend on the diameter of the hole?
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ThudnBlunder
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Re: Hole Through Sphere
« Reply #2 on: Apr 13th, 2003, 5:24pm » |
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Quote:| Wouldn't that depend on the diameter of the hole? |
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Nope. If it did, I wouldn't have posted. 
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| « Last Edit: May 29th, 2004, 7:27pm by ThudnBlunder » |
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aero_guy
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Re: Hole Through Sphere
« Reply #3 on: Apr 13th, 2003, 5:58pm » |
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Well then, this seems like it must be a trick. The question, "What volume of the sphere's material remains?" could be interpreted as, "What volume of the sphere's material still exists?" in which case it would be approximately the original volume, as the material was simply moved, not destroyed. Of course there may be some losses if the material was compressed or burnt or something during drilling.
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ThudnBlunder
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Re: Hole Through Sphere
« Reply #4 on: Apr 13th, 2003, 6:05pm » |
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Easy puzzles shouldn't come with hints. 
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cho
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I've heard this one before, so I'll just give a hint to understand the question; it is the hole that is 6", not the core that was taken out. So if the sphere is 7 or 8 or 12 inches in diameter, the hole must be wide enough so the ring left behind is only 6" thick.
I hope you don't think my hinting your riddle is inappropriate or I'm that stealing your blunder, thud.
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cho
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Oops, I was typing my response before you posted No hints.
I really should register so I can delete these mistake posts.
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ThudnBlunder
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Re: Hole Through Sphere
« Reply #7 on: Apr 13th, 2003, 6:20pm » |
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Quote:| I hope you don't think my hinting your riddle is inappropriate or I'm that stealing your blunder, thud. |
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Not at all. Core? Ring? Thick? Sowing confusion is always welcome. 
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| « Last Edit: Apr 14th, 2003, 4:11am by ThudnBlunder » |
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BNC
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Re: Hole Through Sphere
« Reply #8 on: Apr 14th, 2003, 12:03am » |
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I assume that by '6" long" you mean the length of hole, as measured by a calipper (i.e., ignoring the material that was removed while drilling).
In that case, by going to the extreem case, I would say the volume of a 6" sphere.
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ThudnBlunder
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Re: Hole Through Sphere
« Reply #9 on: Apr 14th, 2003, 2:28am » |
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What is 'a 6" sphere'? In arriving at your answer, what else are you assuming?
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BNC
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Re: Hole Through Sphere
« Reply #10 on: Apr 14th, 2003, 4:46am » |
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Well, I assume that, like you said, the diameter of the hole does not matter. Thus, I take the extreem case of a sphere of 6" diameter and a hole with 0 diameter. In this case, the hole is 6" 'long', and the volume of the sphere is unchanged as compared to the original 6" diameter sphere.
On the other hand, other than the wording of the riddle, I can't (yet) prove this result. 
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ThudnBlunder
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Re: Hole Through Sphere
« Reply #11 on: Apr 14th, 2003, 9:53am » |
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Quite right, BNC. In future, I will give no more hints - no matter how plaintive the pleas. 
PS What does 'extreem' mean? What is a 'calipper'?
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| « Last Edit: Apr 14th, 2003, 9:54am by ThudnBlunder » |
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BNC
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Re: Hole Through Sphere
« Reply #12 on: Apr 14th, 2003, 9:58am » |
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Sorry, m8. English is not my mother tongue, and I do tend to make quite a few errors (both spelling and grammatical). Oh well… maybe some day I’ll learn.
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Icarus
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Re: Hole Through Sphere
« Reply #13 on: Apr 14th, 2003, 4:17pm » |
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Proving it is a fairly simple calculus problem. Try finding the volume of the spherical slice, then remove the volume of the cylinder.
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Kozo Morimoto
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Re: Hole Through Sphere
« Reply #14 on: Apr 16th, 2003, 8:55pm » |
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I still don't get this. The hole is 6" long and it's through the centre. It doesn't say that it comes out the other side, as long as its through the centre (like an 8" sphere with a 6" deep hole). I can't understand how you can solve this without the diameter of the sphere and the diameter of the hole.
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ThudnBlunder
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Re: Hole Through Sphere
« Reply #15 on: Apr 17th, 2003, 2:23am » |
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Quote:| I still don't get this....It doesn't say that it comes out the other side... |
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Yes, the hole goes right through the centre of the sphere and comes out the other side.
Here is the puzzle in verse form:
Old Boniface he took his cheer,
Then he bored a hole through a solid sphere,
Clear through the center, straight and strong,
And the hole was just six inches long.
Now tell me, when the end was gained,
What volume in the sphere remained?
Sounds like I haven't told enough,
But I have, and the answer isn't tough!
Quote:Try finding the volume of the spherical slice, then remove the volume of the cylinder.
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When I first saw this problem in one of Martin Gardner's books, I verified the answer using this method.
Even without deriving the formula for a spherical cap, it is a very tedious calculation.
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| « Last Edit: May 29th, 2004, 7:30pm by ThudnBlunder » |
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LZJ
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Re: Hole Through Sphere
« Reply #16 on: Apr 17th, 2003, 4:07am » |
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Wouldn't the use of solid angles simplify the process?
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Icarus
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Re: Hole Through Sphere
« Reply #17 on: Apr 17th, 2003, 5:48pm » |
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on Apr 17th, 2003, 2:23am, THUDandBLUNDER wrote:
When I first saw this problem in one of Martin Gardner's books, I verified the answer using this method.
Even without deriving the formula for a spherical cap, it is a very tedious calculation. |
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Apparently you did it the hard way:
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SWF
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Re: Hole Through Sphere
« Reply #18 on: Apr 17th, 2003, 6:20pm » |
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If sphere has radius, R, and you take a slice perpendicular to the hole's axis you see a ring of inner radius A and outer radius B. If y is distance of the slice from the center of the sphere:
A2=R2-32 (the 3 comes from being half the hole length)
B2=R2-y2
Area of the cross-section= pi*(B2-A2)=pi*(32-y2)
Volume is based on how this area varies along the 6 inch length of the hole, but notice how the value of R disappeared from the equation. So volume is as BNC suggested after assuming R is irrelevant.
Also each cross section has the same area as a sphere of radius 3...
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LZJ
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Re: Hole Through Sphere
« Reply #19 on: Apr 17th, 2003, 6:23pm » |
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I agree that there's a much simpler way, but I don't fully understand your method, Icarus. From what I see, it is simply taking 2 cones of height 3, radius 3 from a cylinder of radius R and length 6. Perhaps you would care to explain how to use the result to obtain the volume of a spherical slice?
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Icarus
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Re: Hole Through Sphere
« Reply #20 on: Apr 17th, 2003, 7:47pm » |
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Volume is the 1-D integral of area - what I have is the integral of the cross-sectional areas starting from one end of the spherical slice going to the other end. The cross-section at a distance x from the center is a circle of radius (R2 - x2)1/2. The area of these circles is what I am integrating.
To complete the derivation of the formula for the sphere with hole, the cylinder has radius PI(R2 - 32) and length 6, so its volume is 6*PI*R2 - 54*PI.
Taking the difference, the 6*PI*R2 cancels, and you are left with 36*PI, which is indeed the volume of a 6" sphere. (This works for any radius sphere, by the way).
As a sidenote: LJZ - your interpretation of my integral as taking cones from a cylinder happens to be exactly how Archimedes came up with the formula for the volume of a sphere (which he considered the crowning achievement of his life) in the first place. Archimedes noticed that if you add together the cross-sectional area of a sphere of radius R at any distance x from the center and the cross-sectional area of a right-angled cone of base radius R, at a distance x from the vertex, you get the area of a circle of radius R - the cross-section of a cylinder of radius R. On the basis of this, he decided that the volume of a sphere ought to be the difference between the volume of a cylinder of radius R and height 2R (2*PI*R3) and of two cones of base radius R and height R (PI*R3/3 each).
Since the ideas of calculus were far into the future in his time (except the most basic and important: Eudoxus' method of exhaustion, which Archimedes used), he did not consider this to be a proof. So instead, he came up with this really complicated construction using small irregular cones with vertices at the center. This proof clearly required him to know what the formula was before he started it, and for many years scholars wondered how he did it, until in the middle of the 20th century a manuscript was discovered outlining what I described.
So your interpretation of my integral as cylinders and cones was not so mistaken after all!
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LZJ
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Re: Hole Through Sphere
« Reply #21 on: Apr 17th, 2003, 8:38pm » |
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Oh, I geddit, thanks for enlightening me 
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