Hole in a sphere

[bonanova]

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.

[Guest]

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 hole is drilled through a sphere.

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

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

I am still not seeing this. How can the volume of the two spheres below be the same?

[bonanova]

I am still not seeing this. How can the volume of the two spheres below be the same?

Answer this question and you'll understand the puzzle:

How can equal-length cylinders go [completely] through unequal-diameter spheres?

Hint: It's possible to drill a 6-inch-long hole through the earth.

[Guest]

Answer this question and you'll understand the puzzle:

How can equal-length cylinders go [completely] through unequal-diameter spheres?

Hint: It's possible to drill a 6-inch-long hole through the earth.

By not going through the center of the sphere?

[bonanova]

By using a drill with a larger diameter.

[Guest]

So we have spheres of various size. We have drills of various circumference, but all only 6 inches long. I'm still not getting it. How can a six inch long drill with a diameter of any size (1/2 in. or a mile) dirll completely through the earth. I'm picturing a small 1/2 hole and a larger 1 mile depression but still only 6 inches deep.

[bonanova]

So we have spheres of various size. We have drills of various circumference, but all only 6 inches long. I'm still not getting it. How can a six inch long drill with a diameter of any size (1/2 in. or a mile) dirll completely through the earth. I'm picturing a small 1/2 hole and a larger 1 mile depression but still only 6 inches deep.

Visualize the material that's removed from the sphere by the drill.

It comprises two spherical caps sandwiching a right circular cylinder.

Define the length of the hole to be the height of the cylinder.

That's the same as if you climbed inside the hole and measured its length from end to end with a measuring tape.

Note that because the end caps have height of their own, the length of the hole is less than the diameter of the sphere.

Also note that as the diameter of the drill approaches the diameter of the sphere, the length of the cylinder [length of the hole] approaches zero. That is, all the sphere is removed: comprising two hemispherical end caps and no cylinder. This means that any sphere of diameter greater than 6 inches can have a 6 inch hole drilled completely through it.

To understand the puzzle is to understand that it could have been worded worded more helpfully. See next.

But that's part of the puzzle - to understand the condition of "6-inch hole drilled through a sphere", even tho the puzzle is worded [accurately, but] not helpfully:

Drill a hole through a sphere. Make the diameter of the drill large enough that the inside cylindrical surface of the hole measures exactly 6 inches in length. Now calculate the volume of the portion of the sphere that remains = volume of: (sphere - cylinder - 2x [end caps]).

[roolstar]

So we have spheres of various size. We have drills of various circumference, but all only 6 inches long. I'm still not getting it. How can a six inch long drill with a diameter of any size (1/2 in. or a mile) dirll completely through the earth. I'm picturing a small 1/2 hole and a larger 1 mile depression but still only 6 inches deep.

If you look at a doughnut cut through a sphere ( or a regular doughnut), it's height (when you put it on the plate) is about 5 cm.

Now the diameter of the original sphere (before the hole) is actually the diameter of the doughnut itself (the circle) maybe 15 cm!

The same with a hole through earth: The height of the doughnut cut is 6 inch, but the sphere has a much bigger diameter.

And the drill is not 6 inch long at all, the drill can actually be as long as you like but it's width (or radius) has a MAXIMUM* to be calculated in a way to leave a 6 inch doughnut in the sphere in question.

In the case of the earth, its radius has to be a small fraction less than that of the earth.

And in a 3 inch radius sphere, its width has to be 0 and the doughnut's height will then be 6 inch with no hole in the middle!

Of course with a sphere with less than 6 inch diameter, the puzzle doesn't hold.

* Why MAXIMUM? Well, the drill itself can be a regular drill with regular dimensions, but it's going to take a long time to drill through the earth and leave a 6 inch doughnut from the planet, not to mention there will be no place to stand!

Edited February 5, 2008 by roolstar

[Guest]

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 hole is drilled through a sphere.

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

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

I feel like the kid getting off the short bus. However, I am determined to understand this. Where in the original post does it state that the hole will leave only 6 inches left around it? It says "A 6-inch hole". I read that as either a hole 6 inches deep or a hole 6 inches in diameter. Either way, it's not very big and will take away essentially nothing from an object the size of the earth. Now you guys are making it sound as if (from a 2D standpoint) there is a circle with all but six inches cut out. I am seeing something that looks more like a bracelet than a doughnut.

(I do appreciate the patience with the slow kid, too.) Am I just completely misunderstanding the wording of the question?

[roolstar]

I feel like the kid getting off the short bus. However, I am determined to understand this. Where in the original post does it state that the hole will leave only 6 inches left around it? It says "A 6-inch hole". I read that as either a hole 6 inches deep or a hole 6 inches in diameter. Either way, it's not very big and will take away essentially nothing from an object the size of the earth. Now you guys are making it sound as if (from a 2D standpoint) there is a circle with all but six inches cut out. I am seeing something that looks more like a bracelet than a doughnut.

(I do appreciate the patience with the slow kid, too.) Am I just completely misunderstanding the wording of the question?

Not at all Jkyle1980 !

I followed your posts on this forum very closely and know exactly how "SLOW" you are!

You are completely right (at least I think so), I was completely mislead by the same fact as well: It was not so clear in the OP.

But once I read (in an later post) that the cylinder left AFTER the drilling is 6 inch high, I saw the effect of drilling the 6 inch hole through earth more clearly...

And it does look more like a bracelet too!

[bonanova]

I'm certain the question was meant to be a bit demanding.

It was recalled by a friend from an MIT entrance exam.

It requires visualizing the drilled sphere.

Once you see it, drilled, the height of end caps is a non-issue.

The hole has an unambiguous length.

[Guest]

I dropped out of Calc III years ago because I sucked at visualizing three dimensions. I am still just drawing a blank. When you guys are talking about the cylindar, are you meaning what is left of the original sphere or the piece that is removed from the sphere? Because if it is the former, I am just way out of it. (This may also be why I went to Tulsa instead of MIT. Go GOLDEN HURRICANE!!!)

[bonanova]

I dropped out of Calc III years ago because I sucked at visualizing three dimensions. I am still just drawing a blank. When you guys are talking about the cylindar, are you meaning what is left of the original sphere or the piece that is removed from the sphere? Because if it is the former, I am just way out of it. (This may also be why I went to Tulsa instead of MIT. Go GOLDEN HURRICANE!!!)

The cylinder is what is removed from the sphere.

Think of beads on a string.

The string is the cylinder.

[Guest]

Okay, beads on a string. I like that. It's simple. I can see it. Now is the six inches the length of the string or the diameter of the string? Or is it the width of the bead from the hole to the outside?

[bonanova]

Length.

[bonanova]

Length.

[Guest]

So the hole is 6 inches deep? And so how does putting a hole 6 inches deep in the earth make the volume of the remaining earth 36pi?

[Guest]

As long as you told us what the radius of the sphere was then any fool could tell you that your Volume was:

4/3 ¶ r³ - ¶ 6² r ! DUH

[Guest]

As long as you told us what the radius of the sphere was then any fool could tell you that your Volume was:

4/3 ¶ r³ - ¶ 6² r ! DUH

Who is this person that speaks to me as if I needed his (or her) advice?

[bonanova]

As long as you told us what the radius of the sphere was then any fool could tell you that your Volume was:

4/3 ¶ r³ - ¶ 6² r ! DUH

A moment's reflection shows the fool's answer is incorrect.

The answer is 36 pi in**3, independent of sphere radius.

[bonanova]

So the hole is 6 inches deep? And so how does putting a hole 6 inches deep in the earth make the volume of the remaining earth 36pi?

After the hole is drilled completely through the sphere, not just part way into the sphere,

the height[*] of the remaining part of the sphere is 6 inches.

[*] assuming the drill was vertical.

That's what's meant by drilling a 6" [long] hole thru a sphere.

If the sphere was the earth, the drill's diameter must have been just less than the earth's diameter,

leaving a 6" wide bracelet of earth that includes the equator and 3" either side of it.

In the above quote, visualize the drill that went only 6" into the earth.

Now use that drill to go all the way through the earth.

Measure the length of that hole.

If it's not 6", the drill has the wrong diameter.

I'm estimating that sometime, perhaps around Valentine's Day, the sound of a forehead slap will echo through the hole.

[Guest]

I didn't really understand it until this post.. but now I do.

Going back to the calculus answer, in order to find the volume, you'd have the sphere, the cylinder, and two end caps. Adding the cylinder and the end caps gives you the total volume removed from the sphere.

The condition is that the cylinder must be 6 inches long. The diameter is not specified or necessary; it simply must be 6 inches long. Or high. Whichever variable makes you happy. Also, note that this is the cylinder not counting the end caps.

Consider a large sphere and a really, really large person wanting a 6 inch bracelet. You have a drill and various drill bits all infinitely long (or, for all practical purposes, long enough to drill through the sphere with room to spare) but with varying diameter (length isn't important; diameter is). In order to make that 6 inch bracelet, you would have to choose a drill bit with a diameter large enough such that when you drill a hole completely through the sphere (going through the center, of course), the cylinder that you took out (again, not counting the end caps) would be 6 inches long. Thus, when you lay what's left over (the bracelet) flat on a table, it will be 6 inches high.

That is, assuming, that the sphere is large enough to accomplish this.

Consider a sphere infinitesimally larger than 6 inches in diameter (take the limit as diameter approaches 6). If you wanted to drill a hole through the sphere such that the cylinder which is removed (or, alternatively, the bracelet which is left behind) is 6 inches high, well, the sphere is already nearly 6 inches high itself, so you would need an infinitesimally thin drill. At 6 inches exactly, there's no way you can drill a hole which, when calculating the volume removed, would be composed of a 6 inch cylinder and two end caps.

Just my two cents. Hope it helps

[Guest]

Okay, I'm not good at Calculus, but the easiest way i see it is that you throw the remaining part of the sphere into a bucket of water and see how much volume is displaced. However, I'm assuming that this solid sphere has a density greater than water so that it'll sink enough for the volume displacement.

[Guest]

After the hole is drilled completely through the sphere, not just part way into the sphere,

the height[*] of the remaining part of the sphere is 6 inches.

[*] assuming the drill was vertical.

That's what's meant by drilling a 6" [long] hole thru a sphere.

If the sphere was the earth, the drill's diameter must have been just less than the earth's diameter,

leaving a 6" wide bracelet of earth that includes the equator and 3" either side of it.

In the above quote, visualize the drill that went only 6" into the earth.

Now use that drill to go all the way through the earth.

Measure the length of that hole.

If it's not 6", the drill has the wrong diameter.

I'm estimating that sometime, perhaps around Valentine's Day, the sound of a forehead slap will echo through the hole.

So this? (And that sound isn't the slapping of a forehead. It's of the pounding of a forehead against a desk. haha)

[bonanova]

That's the idea...

[roolstar]

That's the idea...

Is it?

Did Jkyle1980 post the right approach in drilling??

I think not!!

According to his picture, the diameter of the hole = diameter of sphere - 6 inch!!

Now that cannot ge right according to the answer and to the posts on this thread!

The 6 inchs in his picture cannot be seen from this angle, they represent the HEIGHT of the remaining part of earth once you put the earth bracelet flat on a plane!!