Sphere Volume

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
The diagram below shows a sphere through which a cylindrical hole has been drilled straight through its centre (top to bottom).

After drilling, the measured distance shown is 6 inches; what is the volume of material remaining in the sphere?

For clarification, the 6 inch distance is the distance over which the full diameter of the drill was cutting the sphere material – with the ‘north’ & ‘south’ pole domes removed in the drilling process. The diagram below is for clarification of the question, no dimension other than the 6 inches can be inferred from the diagram.

This is not a trick question, there is sufficient information given above to answer the question.

drilled sphere.jpg
 

wayneh

Joined Sep 9, 2010
17,496
This is not a trick question, there is sufficient information given above to answer the question.
If that's true, then the answer is immediately available at the limit where the drill diameter approaches zero and the diameter of the sphere approaches 6 inches.

Proving that it's true is the more challenging question.
 

WBahn

Joined Mar 31, 2012
29,979
The diagram below shows a sphere through which a cylindrical hole has been drilled straight through its centre (top to bottom).

After drilling, the measured distance shown is 6 inches; what is the volume of material remaining in the sphere?

For clarification, the 6 inch distance is the distance over which the full diameter of the drill was cutting the sphere material – with the ‘north’ & ‘south’ pole domes removed in the drilling process. The diagram below is for clarification of the question, no dimension other than the 6 inches can be inferred from the diagram.

This is not a trick question, there is sufficient information given above to answer the question.

View attachment 161469
If that's true, then it must be independent of the diameter of the hole, meaning that we can pick ANY convenient diameter. So let's look at the two extreme cases.

1) The diameter of the hole is zero. This means that the entire sphere is left and the volume remaining is 4/3 pi R^3 where R = 6 in/2 = 3 in, volume = 36π in^3.

However, it must also be true in the case where the diameter of the sphere is large and the diameter of the hole approaches the diameter of the sphere. Unfortunately, I don't see any obvious simplification in this case.

But we can determine the answer through direct integration.

V = ∫ (πr²(y) - πx²(y)) dy evaluated between y = -3 in and y = +3 in.

Where the integrand is the area of the annular ring at height y between the drilled hole and the surface of the sphere.

Due to symmetry, this can be reduced to

V = 2π ∫ (r²(y) - x²(y)) dy evaluated between y = 0 and y = +3 in.

r(y) is the radius of the circle formed by the intersection of the sphere and the plane at height y.

x(y) is the radius of the circle formed by the intersection of the hole drilled in the sphere and the plane at height y.

r(y) = √(R² - y²)

x(y) = √(R² - (3 in)²)

V = 2π ∫ ( (√(R² - y²))² - (√(R² - (3 in)²))² ) dy

V = 2π ∫ ( R² - y² - R² + 9 in² ) dy

V = 2π ∫ (9 in² - y²) dy

V = 2π (9 in²·y - (1/3)·y^3) | (0 to 3)

V = 2π (27 in^3 - (1/3)·27 in^3)

V = 2π (27 - 9) in^3

V = 2π (18) in^3

V = 36π in^3
 

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
What is interesting about this question is that the resultant remaining sphere volume is constant for any sphere diameter.

wayneh gave the correct solution without resorting to calculus or other mathematical proofs, and WBhan proved the answer with calculus.

It is also solvable by subtracting the two pole volumes and the 6 inch drilled volume from the original sphere volume (with the terms cancelling out to give a solution of 36.π cubic inches).
 

WBahn

Joined Mar 31, 2012
29,979
What is interesting about this question is that the resultant remaining sphere volume is constant for any sphere diameter.
Yes, it's interesting, but hardly unfathomable. There are many problems that behave like this and often they can be discovered by looking at differential relationships. The classic example is asking how far off the ground would a circular string be if it resulted from adding one meter to a string that was wrapped around a perfectly spherical Earth at the equator. The answer is independent of the radius of the Earth because the derivative of the circumference of a circle with respect to the radius is independent of the radius itself, only the change in the radius.

wayneh gave the correct solution without resorting to calculus or other mathematical proofs, and WBhan proved the answer with calculus.
Actually, he did resort to "other mathematical proofs" of a sort -- he started out by stating, "If that's true, then...." In other words, he explicitly tied the validity of his solution to the validity of your claim, correctly refusing to stake any claim as to the validity of either since neither had been proven adequately. He was merely stating a required consequence of your assertion.

It is also solvable by subtracting the two pole volumes and the 6 inch drilled volume from the original sphere volume (with the terms cancelling out to give a solution of 36.π cubic inches).
Of course you can subtract the volume of the two poles and the volume of the cylinder to get the answer -- but how do you find the volume of one of the poles? I would use integral calculus to do it since I don't see any way to approach it via geometry or trigonometry (which is not to say that this is impossible).
 
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Thread Starter

Hymie

Joined Mar 30, 2018
1,277
Of course you can subtract the volume of the two poles and the volume of the cylinder to get the answer -- but how do you find the volume of one of the poles? I would use integral calculus to do it since I don't see any way to approach it via geometry or trigonometry (which is not to say that this is impossible).
Based on the diagram below:-

The volume of the sphere = 4/3.π.R˄3

The volume of the dome cap = ((π.h˄2)/3).(3R-h) where h = R-3

The volume of the drilled section = π(R˄2-9) x 6

Using the above equations, everything cancels out to give 36. π

Sphere d.jpg
 

WBahn

Joined Mar 31, 2012
29,979
Based on the diagram below:-

The volume of the sphere = 4/3.π.R˄3

The volume of the dome cap = ((π.h˄2)/3).(3R-h) where h = R-3

The volume of the drilled section = π(R˄2-9) x 6

Using the above equations, everything cancels out to give 36. π

View attachment 161590
And where does

The volume of the dome cap = ((π.h˄2)/3).(3R-h)

come from?

I'm pretty sure that far, far fewer than 1% of the people that could give you the formula for both the volume of a sphere and the surface area of a sphere could spit out the formula for the dome cap of a sphere. I also suspect that finding this formula, even on the internet, is not the simplest thing, though I'm sure it can be found without too much effort.

But, if going and looking up the formula for the dome cap and then just using it without showing how it is arrived at is fair game, then why isn't just going out to any of the many sites that have the solution to this problem and posting the solution without any justification at all not fair game?
 

Thread Starter

Hymie

Joined Mar 30, 2018
1,277
The question intrigued me in that it appears there is insufficient information to give a definitive answer (it may similarly intrigue others).

To that end, I would have posted the question regardless of how complex the maths required to reach the solution.
 

MrAl

Joined Jun 17, 2014
11,396
Hi,

Sort of interesting :)

Next, look at different ways to develop the equation for the circle at the northern pole formed by the drilling operation given the same input data.
 

wayneh

Joined Sep 9, 2010
17,496
But we can determine the answer through direct integration.
There was a time I could have done that quickly enough to be worthwhile. For a quick answer to an internet post, I'm sad to say it would have taken me too long to get back in the integration groove. :(
 

cmartinez

Joined Jan 17, 2007
8,220
There was a time I could have done that quickly enough to be worthwhile. For a quick answer to an internet post, I'm sad to say it would have taken me too long to get back in the integration groove. :(
Same here ... although I perfectly understand the answer, and can easily follow the logic involved, when it comes to actually grabbing a pencil and paper and performing the calculations involved I'm afraid to say that my brain is in dire need of a few splashes and squirts of WD-40
 

WBahn

Joined Mar 31, 2012
29,979
There was a time I could have done that quickly enough to be worthwhile. For a quick answer to an internet post, I'm sad to say it would have taken me too long to get back in the integration groove. :(
I was very, very fortunate that when I was learning this stuff all of my teachers, both in high school and college, taught it in a way where the fundamental concepts were paramount and the mechanical details secondary -- it is often taught exactly the other way around. Since this is a good match for the way I tend to internalize things, most of the basics stuck very well so that, even after 30+ years of separation, I can still go back to the basics and build up from there without having to look anything up.
 

WBahn

Joined Mar 31, 2012
29,979
So can I ... but the experience is akin to trying to stand up with both legs numb after too long a time spent sitting on a chair ... :eek::confused:
And therein lies a big difference -- I actually enjoy doing it. I don't know that that necessarily says positive things about me, though.

Every few years (though not since I've gotten married, which probably says something about my free time before and after) I've actually gone through and tried to walk through building up a table of integrals and derivatives from the fundamental definitions to see how far I could get before having to pull out a calc book to find the trick needed to do something. I could usually make it through all of the usual suspects including the trig and inverse trig functions. A friend and I, when studying for the PhD qualifier, got off on a side tangent and spent several hours developing, from scratch, the table of Laplace Transforms (forward and inverse). Again, it doesn't necessarily say good things about us that we actually had a lot of fun doing it (and this was before I got married, though I'd probably get similarly sidetracked today pretty easily).
 

wayneh

Joined Sep 9, 2010
17,496
And therein lies a big difference -- I actually enjoy doing it.
Yeah, I enjoy it too and I suspect that @cmartinez does too. But that doesn't mean it's quick and easy enough for us to justify sitting with a cup of coffee and taking the time to work through it, however enjoyable that might be. If I was building something and it was part of a project, I wouldn't hesitate. But just to answer a post on a forum? I just don't have the energy!

Speaking of things I enjoy that are becoming harder and slower, I'm heading out for a 5K run. Beautiful fall leaves here now.
 
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