Posted 21 Aug 2007 Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute. She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way. A 6-inch [long] hole is drilled through [the center of] a sphere. What is the volume of the remaining portion of the sphere? The hard way involves calculus. The easy way uses logic. 0 Share this post Link to post Share on other sites

0 Posted 22 Oct 2010 The point is that the answer does not depend on the shpere's diameter. Start with sphere larger than 6" diameter, where it's easy to drill the hole but hard to compute the remaining volume. Reduce the sphere's size; the hole's diameter shrinks, and the remaining volume gets close to the original volume. When you get down to exactly 6" diameter, the hole has vanished, and the remaining volume is easy to calculate. It's precisely the original volume: [4/3]pi*[3in]**3 = 36 pi cu in. See the bold part. VANISHED implies the hole is not there. Hence, wrong. 0 Share this post Link to post Share on other sites

0 Posted 22 Oct 2010 it's just the volume of the cylinder pi*r^2*"6 -4/3*pi*r^3 0 Share this post Link to post Share on other sites

0 Posted 22 Oct 2010 What if the sphere already had a hole (of the same size of the drill) and you just re-drilled through it? Then the volume of sphere would remain same as before. 0 Share this post Link to post Share on other sites

0 Posted 24 Oct 2010 What if the sphere already had a hole (of the same size of the drill) and you just re-drilled through it? Then the volume of sphere would remain same as before. I guess the answer to that would be a sphere with a hole is not a sphere. 0 Share this post Link to post Share on other sites

0 Posted 24 Oct 2010 See the bold part. VANISHED implies the hole is not there. Hence, wrong. Well, one could say vanishingly small and remove this objection. For a short, tongue in cheek treatise on the existence of things with attributes of zero, 0 Share this post Link to post Share on other sites

0 Posted 27 Oct 2010 I'd like to believe it's 4/3 X 3.14 X 3 X 3 X 3 cubic inches which is the result when the hole's diameter is 0. 0 Share this post Link to post Share on other sites

0 Posted 27 Oct 2010 if you drilled a hole through a round "speaker" there would no longer be any "volume" .....how far out of the box am I?? 0 Share this post Link to post Share on other sites

0 Posted 28 Oct 2010 if you drilled a hole through a round "speaker" there would no longer be any "volume" .....how far out of the box am I?? If the speaker were a sphere, sure ... good one. 0 Share this post Link to post Share on other sites

0 Posted 30 Nov 2010 I'd like to believe it's 4/3 X 3.14 X 3 X 3 X 3 cubic inches which is the result when the hole's diameter is 0. Assuming (or guessing) that the volume remaining will be the same no matter how large or small the sphere is, then we know that as the diameter of the sphere approaches 6", the radius of the hole approaches 0" and the remaining volume is 36pi. 0 Share this post Link to post Share on other sites

0 Posted 16 Mar 2011 (edited) the answer is the volume of a sphere with a diameter of 6 (I can't remember the formula for spherical volume) my logic: if the remaining volume of the sphere is the same no matter the diameter of the cylinder, reduce the cylinder to a line with a length of 6 inches, the remaining volume is the sphere with a 6 inch diameter Edited 16 Mar 2011 by b1soul 0 Share this post Link to post Share on other sites

0 Posted 3 May 2011 Don't read unless you're sure you want to.: D The answer is zero, because the sphere would've been aired up, unless wooden, which is highly unlikely. The answer is zero, because the hole would "pop" the sphere and all the air would be gone. 0 Share this post Link to post Share on other sites

0 Posted 3 May 2011 its the normal length of a sphere with 6 as a diamator logic: with a flat side the diamator of the cylandir could be the diamaotr of the largest circle that can fit on the side but a sphere has no flat sides so the cylandir can not have a diamator 0 Share this post Link to post Share on other sites

0 Posted 12 May 2011 The problem is people are thinking of a drill bit and the distance the 'point' travels. For a sphere the size of the earth they are envisaging a drill (of any size but lets say 1mile diameter) and seeing the point touching the north pole and drilling down and exiting the south pole - which is an 8000mile distance. Think instead of a hole saw (google image hole saw if you dont know what one is - imaging one is essential to the problem) - it isnt a 'solid' drill bit - it is only a perimeter. Now think of a hole saw almost the size of the earth such that when you lower it down over the earth it only cuts through 6 inches of material. You can see that the smaller the hole saw the more material it would cut through - if your hole saw was 1mile wide it would bore through all 8000 miles... if your hole saw is 4000 miles in diamater it cuts through less material. if your hole saw is 8000 miles wide less 6 inches it only bores through a minute amount of material. This is the core of the problem - people think the length of a hole (that passes exactly through the center of the sphere) is measured from the two most distant points - ie north pole to south pole when it isnt - it is the shortest distance - if you could stand on that huge ring left after hole sawing the earth the 'thickness' as you measure down is 6 inches. 0 Share this post Link to post Share on other sites

0 Posted 18 Oct 2011 Hello, thank you for the awesome puzzle! Now, having read a lot of the discussion, I have come to think that there are actually 3 answers possible: 1,2) If the length of the hole is equal to the diameter of the sphere (6 inch hole, 6 inch diameter sphere) then the answer is either 36pi or 0. That is because in order to drill such a hole in such a sphere the diameter of the drill must be either 0 inches or 6 inches (again must think about the length of the INSIDE of the sphere after the hole's been drilled out. Like standing inside the hole - if there's anything visible left, then the hole is less than 6 inches). So, if the diameter of the drill is 0 inches, the remaining volume of the sphere equals to the original volume of the sphere (36pi), as nothing was drilled away. If the diameter of the drill is 6 inches, the sphere is, of course, completely drilled away with nothing left, so you can only imagine the hole, and the volume is thus 0. 3) If the length of the hole is smaller than the diameter of the sphere (6 inch hole, 7 inch diameter sphere) then the answer is impossible to figure out logically, without computing it. And if the problem would have no answer that is possible to figure out logically, thus it would have no answer at all, given it is a logic puzzle. So actually, there's three answers to this problem, given the aforementioned reasoning - 36pi, 0 and No Answer. Please, correct me if my reasoning somewhere is wrong. p.s. Actually, while writing this post I have lost my line of thinking several times, and rewrote it accordingly =p This is the final version that I can finally read without getting confused myself... Made my night *sleepy*. 0 Share this post Link to post Share on other sites

0 Posted 14 Dec 2011 Since it (almost) does not matter what size the sphere is, just use the easiest one. A six inch shere would be nice. Then the hole would be zero diameter and all the sphere would be the remainder. Use the formula 4/3 pi r^3 with a radius of 3 inches. 0 Share this post Link to post Share on other sites

0 Posted 15 Dec 2012 IT SAID DRILLED THROUGH SO THE AMOUNT REMAINING IS 0 0 Share this post Link to post Share on other sites

Posted

Maybe this has already been posted. A friend asked me this a while back, and I answered her in less than a minute.

She said I was a genius. But I said there were two ways to arrive at the answer, and I simply chose the easier way.

A 6-inch [long] hole is drilled through [the center of] a sphere.

What is the volume of the remaining portion of the sphere?

The hard way involves calculus. The easy way uses logic.

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