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# Hole in a sphere

Go to solution Solved by bonanova,

## Question

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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• Solution
The volume of the spherical caps is given by:

where

[*] h = the height of the cap (difference between r and the distance from the centre of the[/*:m:1cc31]

sphere to the centre of the circular end of the hole)

Kudos to cpotting for the cap formula.

Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.

Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L.

Swizzling cpotting's cap formula, V[cap] = pi/3 (2R*R*R - 3R*R*L + L*L*L)

Cylinders are ho-hum, V[cyl] = 2pi*L*r*r = (2pi/3) (3R*R*L - 3L*L*L)

V[removed by drilling] = V[cylinder] + 2V[cap]

doing the math,

V[removed] = (4pi/3)R*R*R - (4pi/3)L*L*L

Pretty amazing: the volume removed by a hole of length 2L is the difference of the volumes of two spheres: one of radius R, the other of radius L.

So the remaining volume is simply the volume of a shpere with radius L. [hint-hint at the logical solution]

V[remainder] = (4pi/3)L*L*L = 36pi.

My friend wouldn't have posed a math problem [boring] and she wouldn't have left out critical information. Therefore the answer couldn't depend on the radius of the sphere. I chose a sphere size [radius=3] that would make 0 volume removed [a hole of length 6 and diameter 0]. With nothing removed, the remaining volume is the original volume: (4pi/3) 3*3*3.

It took me about 15 seconds to say 36pi, and the moment [if not the girl!] was mine.

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• 0 Well,I must not be a genius because the easy way is not jumping in my head. The difficult way would be to determine the volume of the cut out section and deduct it from the volume of the sphere. I am too lazy to do all the work actually involved with doing that though.

It does seem to me that it would be fairly important to know the orientation of the hole. It is natural to assume the hole goes directly through the center of the sphere, but that is really not stated in the question. If I am not mistaken, the problem is a lot more difficult if the hole is off center.

Please post at least the easy answer. I really want to know if I should kick myself for not seeing it immediately.

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• 0 I'm going out on a limb, saying it's more of a lateral thinking puzzle and the volume is none(?). If you drill a whole though a sphere then it is no longer a sphere (at least in the technical sense that i learned back in geometry).

We need not to be let alone. We need to be really bothered once in a while. How long is it since you were really bothered? About something important, about something real?

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• 0 IMO, riddles are more fun when posters get to mull over solutions without the OP providing an answer so soon after posting the riddle. But this is bonanova's riddle and he can do as he pleases; I'm just adding my 2 cents.

I'm going out on a limb, saying it's more of a lateral thinking puzzle and the volume is none(?). If you drill a whole though a sphere then it is no longer a sphere (at least in the technical sense that i learned back in geometry).

No, it's not a lateral thinking puzzle and it doesn't matter if a sphere with a hole can still be called a sphere or not.

To clarify, I believe bonanova is speaking of a hole 6 inches long of an unspecified diameter. Am I correct bonanova?

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• 0 If you take the meaning of volume of the sphere as the amount of space it takes up then its volume would be the same. If you take volume as the amount of material that makes up the sphere I have no idea

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• 0

Clarifications:

 the hole is a circular cylinder of empty space whose axis passes through the center of the sphere - just as a drill would make if you aimed the center of the drill at the center of the sphere and made sure you drilled all the way through.

 the length of the hole [6 inches] is the height of the cylinder that forms the inside surface once the hole is drilled. picture the inside surface as viewed from inside the hole and measure the length of that surface in the direction of the axis of the drill.

in this sense, you could for example drill a 6-inch hole through the earth. the diameter of the hole would be huge, and you'd just have a tiny remnant of the earth left. but if you could set it on a table [a big table] it would be 6 inches high.

you of course could not drill a 6-inch hole through a sphere whose diameter was less than 6 inches. it was actually this fact that led me to the logical answer and made me a genius for a couple of minutes.

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Well,I must not be a genius because the easy way is not jumping in my head. The difficult way would be to determine the volume of the cut out section and deduct it from the volume of the sphere. I am too lazy to do all the work actually involved with doing that though.

I did the calculus afterward, and however you do the calculation, it's difficult. The remaining volume is a volume of revolution with requires finding the cross sectional area and spinning it thru 360 degrees. It's no easier to compute the cylindrical volume removed, cuz there are spherical "caps" on the cylinder which don't have formulas that I could find.

The logical way is easier, and I'll post it in a day or so if you don't get it... have fun.

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• 0 It's no easier to compute the cylindrical volume removed, cuz there are spherical "caps" on the cylinder which don't have formulas that I could find.

The volume of the spherical caps is given by:

where

[*] h = the height of the cap (difference between r and the distance from the centre of the[/*:m:79969]

sphere to the centre of the circular end of the hole)

This still doesn't make it easy to calculate

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• 0 I may be missing something here, as I was a Literature Major not a Math guy, but wouldn't you need 1 more piece of info regardless of semantics?

If you mean by "a six inch hole through a sphere" a six inch long hole "through" the sphere(all the way), then you need to know the diameter of the hole. We know the diameter of the sphere is 6 inches plus pi * h*h * (r - h / 3) , but the hole could be from microscopic to almost 6 inches plus pi * h*h * (r - h / 3) in diameter; thus, size matters.

If you mean by "a six inch hole through a sphere" a six inch diameter hole through a sphere of unknown size, then you need to know the diameter of the sphere. It can be from a little over 6 inches to some outrageously large finite number; thus, size matters.

The moral of the story is: your girlfriend lied. Size DOES matter. But seriously, if the clarification above, in fact, means that it's a 6 inch hole THROUGH a sphere, we know only that the sphere must be 6 inches plus pi * h*h * (r - h / 3) in diameter, or one of two things are incorrect: either the hole is NOT 6 inches long, or it's not THROUGH the sphere - it is merely INTO the sphere. But we still don't know the diameter of the hole. IF it is say 5.9" in diameter, then we've made a really cool hat and can figure out the volume. If it's, say, 1" in diameter then we have a giant bead on our hands and can figure out the volume. But without this extra piece of knowledge, the answer set, HAS to be somewhere between > 0 and 4/3 pi times (6 inches plus pi * h*h * (r - h / 3)) cubed.

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The moral of the story is: your girlfriend lied. Size DOES matter. Writersblock,

Still laughing ... I love it!

My only reply is ... ok I have two replies ...

 she never complained and

 the [apparently lacking] size spec is an important part of the logical solution.

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• 0 Ok, so I still didn't get this so I spent like 4 1/2 hours doing some math. I think I get what you are saying now. The height of the "caps" depend entirely upon the radius of the sphere and the radius of the hole, and thus, so does the removed volume. I am still not convinced you are 100% right, but

I can see that because the length of the hole must always = 6, if you shrink the radius of the sphere = to the length of the hole, and then shrink the radius of the hole = 0, then the volume that is left will always be left, regardless of the size of the hole. That leaves us then with a sphere with a radius of 3, where your 36pi works. The only thing that changes is the "thickness" of the remaning donut, which at some point on the upper sizes would reach theoretical limits of thinness. Like if you have the earth drilled out so that only a 6 inch band remains. (Yes, I know it's not a perfect sphere but it suffices for an example of huge.

Again, my math is weak, so I can't prove this, but I think I smell what you are cooking logically.

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• 0 Wow. I haven't felt this dense since first year law school. I really need to bone up on my math skills. Here's what I did, and I capitulate that 36pi must be correct and bow to bonanova's quick mind.

My math - tell me if I screwed up anywhere.

Sphere volume = 4/3P(R*R*R) where P = pi and R = radius of the sphere

Cylinder Volume = P(r*r)L where P = pi and r = radius of the hole and L = length of the hole

Spherical dome Volume (Cap) = ((P(h*h)*(3R-h))/3) where P = pi and h = height of the cap and R = radius of the sphere [taken from http://www.monolithic.com/construction/formulas.pdf]

so the remaining volume = 4/3P(R*R*R) - (P(r*r)L + (2 ((P(h*h)*(3R-h))/3)))

Define h

2 caps height plus L = 2R so

2h+L=2R where L=6 so

2h+6 = 2R

2h= 2R-6

h= ?*(2R-6)

Define R

the original R must always be 1/2 L+h

If the hole length is 6 inches and goes through a sphere, then

R = 1/2 6 + h. so

R= 3+h

Define R in terms of h

h= ?*(2(3+h)-6)

h= ? * (6+2h-6)

h=h WTF? Infinite?

Try again

R=3+ (1/2*(2R-6)

R=3+R-3

R=R WTF? Infinite?

R-h=3

3+h-h=3

3=3 Infinite

So, regardless of

volume = 4/3P(R*R*R) - (P(r*r)L + (2 ((P((1/2(R-L))*(1/2(R-L)))*(3R-(1/2(R-L))))/3)))

I know that the relationship between R and h are dependant for an infinite set of numbers when L=6. Therefore if h=0 then R=L.

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• 0 Here's the mathematical solution:

Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.

Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L...

I intuitively knew what the answer would be, but I have been struggling to come up with the mathematical proof of it. Looking at your math, though, I am puzzled. How do reason that the area of hole's cross-section (r*r) equals the area of the sphere's cross-section (R*R) less the the square of half the hole's length?

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• 0

Here's the mathematical solution:

Radius of sphere = R, radius of hole = r, length of hole = 2L [so L=3], height of cap = h.

Since r*r = R*R - L*L and h = R-L, we can eliminate r and h and do everything in terms of R and L...

I intuitively knew what the answer would be, but I have been struggling to come up with the mathematical proof of it. Looking at your math, though, I am puzzled. How do reason that the area of hole's cross-section (r*r) equals the area of the sphere's cross-section (R*R) less the the square of half the hole's length?

R*R = r*r + L*L because you can draw a right triangle where R is the hypotenuse [Pythagorus].

Hint: slice the thing along the hole's axis running vertically.

Go from the sphere's center horizontally to the surface of the hole - that's a distance r.

Go straight up to the top of the hole - that's a distance L [at right angles to r]

Go back to the center of the sphere - that's a distance R and is the hypotenuse.

When I answered her question, I hadn't proved 36pi is true for all spheres.

I just guessed. [it's really a quite surprising result!] But, for all she knew

I did all this math in my head in 15 seconds! We were colleagues, and I would

place her IQ somewhere north of 160 - high enough to think it could be

done, and ... high enough that she usually left me in the dust in our work.

She was impressed. It was an moment to savor, and I thought I'd share the story.

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• 0 Well thanks for sharing. I enjoyed crunching my brain on it.

IF anyone is interested, I think the easiest way to think about it is this:

L is the length of the cylinder and is a constant at 6" and R is the radius of the sphere, r is the radius of the cylinder's head, h is the height of the dome on top and bottom of the cylinder that is drilled out. Ok, we are defined.

L must stay constant, so as R expands toward infinity, r and h, and thus the volume of the cylinder and the 2 domes, must expand proportionatly so as to keep the original shape a sphere, and to keep L constant. Thus, for every sphere, where L is constant at 6, the remaining volume is the volume of a sphere where L = 2R. ##### Share on other sites
• 0 I have been looking at this one for two days. If you drill a 6 inch hole through a 6 inch sphere, well you just cant do it. Right?

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• 0 Okay - I've been at this for a while. Thought I would try to find a way to show that the volume of the remainder of the sphere is constant. This would prove the "intuitive" answer that the volume is equal to a sphere 6" in diameter. However, I seem to have proven the opposite . This is only the third time in 27 years that I have used my high school calculus, and I was hoping someone out there may see my error and correct me.

p = pi

D = the diameter of the sphere

d = the diameter of the hole

R = the radius of the sphere (.5D)

r = the radius of the hole (.5d)

H = the height of the hole (end to end)

h = the height of the hole (centre to end = .5H)

After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.

The formula for a barrel with sides bent to the arc of a circle

= pH(2DD + dd) / 12

= (1/12)pH(8RR + 4rr)

The formula for the cylindrical portion that was removed

Therefore the formula for the volume of the ring shape is

and since RR = rr + hh (thanks bonanova - I see that now), we have

which expands to

Now, if we take the derivative of this function, we should see the rate of change in volume with respect to the radius of the hole. This should be 0, meaning that the volume remains constant. Unfortunately, that is not what I am getting.

rr

(constants have been bolded including H and h [which = 6 and 3])

dx®/dy = 2r + 0 + 2r - 2r

= 2r

This means the volume varies with the radius of hole. That would mean that the volume is not always equal.

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• 0
p = pi

D = the diameter of the sphere

d = the diameter of the hole

R = the radius of the sphere (.5D)

r = the radius of the hole (.5d)

H = the height of the hole (end to end)

h = the height of the hole (centre to end = .5H)

After drilling, the remaining ring shape is equivalent to a barrel of unknown width (D), 6" high (H) and with the central cylinder portion removed.

The formula for a barrel with sides bent to the arc of a circle

= pH(2DD + dd) / 12

= (1/12)pH(8RR + 4rr)

cpotting,

kudos for finding all these neat [cap and barrel] formulas.

That's does all of the calculus work.

I found it useful to put everything in terms of R and h.

You can do this by noting that RR = rr + hh [Pythagorus]

V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR - 4hh]/6 = 2ph [RR - hh/3]

V[cylinder] = height x area = Hprr = 2ph [RR - hh]

V[barrel] - V[cylinder] = 2ph [RR - hh/3 - RR + hh] = 2ph [2hh/3] = 4phhh/3.

Recalling that h=3,

V[barrel] - V[cylinder] = 36p [a constant].

Looking at your derivation, everything is correct.

you'll see that all the rr terms add up to zero.

Your derivative expression is correct, also,

except that you should include the constants (1/12)pH8, etc...

[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]

You'll see again that the r-dependent terms add up to zero,

and the derivative is zero.

OK?

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• 0 You're all making this too hard. You just need to know the area of each slice of the donut, which is simply the difference of two circles.

L = length of bore

x = radius of sphere cross-section at height z

A = area of donut cross-section at height z

Forming two simple right triangles:

r? = R? - L?/4

x? = R? - z?

Using the formula for the area of a circle

A = ?(x?-r?) = ?(L?/4-z?)

The area of the cross is independent of R, so the full volume is constant.

To find the volume, you can integrate the area from z=-L/2 to L/2, but it is simpler to use the degenerate case.

R=3, r=0

V = 4?R?/3 = 36? cu.in.

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• 0

Very nice!

There are two steps to get to 36pi.

 deducing it doesn't depend on R.

 computing the R=3 case.

Step  can be done logically or mathematically.

Wordblind absolutely wins the prize for the slickest math.

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• 0 cpotting,

kudos for finding all these neat [cap and barrel] formulas.

That's does all of the calculus work.

I found it useful to put everything in terms of R and h.

You can do this by noting that RR = rr + hh [Pythagorus]

V[barrel] = ph [8RR + 4rr] /12 = ph [8RR + 4RR - 4hh]/6 = 2ph [RR - hh/3]

V[cylinder] = height x area = Hprr = 2ph [RR - hh]

V[barrel] - V[cylinder] = 2ph [RR - hh/3 - RR + hh] = 2ph [2hh/3] = 4phhh/3.

Recalling that h=3,

V[barrel] - V[cylinder] = 36p [a constant].

Looking at your derivation, everything is correct.

you'll see that all the rr terms add up to zero.

Your derivative expression is correct, also,

except that you should include the constants (1/12)pH8, etc...

[if y(x) = Cxx, then dy/dx = 2Cx, not just 2x]

You'll see again that the r-dependent terms add up to zero,

and the derivative is zero.

OK?

Great. Thanks bonanova - I see where I went wrong now.

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• 0

So you're saying that, if you are a Doughnut baker, making the hole bigger while maintaining the same height will not save you any money...

But still your customers will intuitively feel you're cheap so they'll switch to another bakery....

And the only customers you may keep are the guys on this forum.....

Lovin' it! Edited by roolstar
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• 0 Im sorry, I still fail to grasp this logical concept. if you drill a hole with a diameter of 0, then have you really drilled a hole? if so you must have done it with an invisible drill bit. Then again i could be missing the point entirely

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Im sorry, I still fail to grasp this logical concept. if you drill a hole with a diameter of 0, then have you really drilled a hole? if so you must have done it with an invisible drill bit. Then again i could be missing the point entirely
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.

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