Volume of a Sphere | Hacker News Reader

Volume of a Sphere

solipsys.co.uk

62 pointsColinWright40 comments

Generalizations to higher dimensions[1] also have some fun properties. For one, the volume of a n-dimensional sphere (ball) goes to zero as n goes to infinity.

But not in a trivial way: given a radius r, you can find the number of dimensions n that maximizes the volume of the n-sphere. For r=1 it's 5 dimensions, but for r=3 it's 55 dimensions.

Numberphile[2] has a video on higher dimensional spheres which talks about this weirdness.

[1]: https://en.wikipedia.org/wiki/Volume_of_an_n-ball

[2]: https://www.youtube.com/watch?v=mceaM2_zQd8

> For one, the volume of a n-dimensional sphere (ball) goes to zero as n goes to infinity

Just to provide extra clarity: for a fixed radius r, the volume goes to zero as n goes to infinity

Yes important detail, was a bit sloppy there, thanks.

I am going to sugest something about that analysis is flawed as r=1 is unit less. Suggesting measurements in feet and nD feet would favor 55d, but yards / nD yard would favor 3D.

Though I just woke up so I may be missing something obvious.

The idea is to measure relative sizes. So if your first sphere has radius r and the second sphere has radius R, where R = 3r, you can then reason about which n maximizes volume for each. Remember that volume is going to be in units of r (and R) to the power of n.

For unitless numbers "9 > 1", but clearly "9 inches > 1 foot" is not correct.

Units are just conversion factors. They can never change the actual result.

A 5-dim yard and a 6-dim yard convert to a different number of n-dim feet (243 vs 729).

- A 5-d sphere with R=1yd would have volume of ~5.264 yd^5 or 1279 ft^5.

- A 6-d sphere with R=1yd would have volume of ~5.168 yd^6 or 3767 ft^6.

> Generalizations to higher dimensions[1] also have some fun properties. For one, the volume of a n-dimensional sphere (ball) goes to zero as n goes to infinity.

Well, the same thing happens with cubes, doesn't it? (as long as the side is smaller than 1)

From what I gather, the difference is that it happens for n-dimensional spheres regardless of the radius, while as you say for n-dimensional cubes only for those with sides less than one. But I'm no expert so.

Is volume defined for a higher dimension object? I mean, is it like the surface area of a sphere is to the are of a circle? Or as the surface area of a sphere is to the circumference of a circle?

Yes, although sometimes it's referred to as "hyper-volume".

But one dimensional objects have length, two dimensional objects have area, three dimensional objects have volume, and higher dimensional objects all have their equivalent concept.

Nice post. The graph of the sphere, the cone and the cylinder probably needs a 3D version just before it, so it's clear that they are 3D objects. (With a small doll or plastic eye to show the point of view?)

[Request for someone with a lot of Blender capabilities: Make a 3D version of the graph, where the back half of the objects are opaque and the front half are transparent. Since you already have the 3D model, bonus points for another version where the top third of the objects is transparent and the bottom two thirds are opaque.]

> Nice post.

Thank you.

> The graph of the sphere, the cone and the cylinder probably needs a 3D version just before it, so it's clear that they are 3D objects.

Sadly, my skills aren't up to that, and I don't currently have time to acquire them 8-(

> Request for someone with a lot of Blender capabilities: Make a 3D version of the graph, where the back half of the objects are opaque and the front half are transparent.

That would be cool.

> ... bonus points for another version where the top third of the objects is transparent and the bottom two thirds are opaque.

Even cooler.

Sadly, the post is sinking without trace, so I doubt anyone will see the request.

As an aside, if you take the area of a circle, A = pi r^2, and take the derivative wrt r, you get the circumference 2 pi r. Similarly, the derivative of the volume of a sphere, V = 4/3 pi r^3, is the surface area A = 4 pi r^2.

It's sort of obvious, when you think about it, but it wasn't obvious to me before thinking about it :-)

Isn't it the case that integral of surface area is volume, that seems more obvious to me. The derivative happens to be reverse of integral and hence area happens to be derivative of volume.

Yeah, back when I was first introduced to calculus and integration, I was surprised to see ∫4pir^2 dr = 4pi ∫r^2 dr = 4pi*r^3/3. Then I learnt about the Cavalieri's principle and it comes clear in mind.

It's cool ... I mention that in a different (but related) post here:

https://www.solipsys.co.uk/new/ConsideringASphere.html#toc_n...

> It was only 282 years ago that Euler presented in his textbooks the exact formula for the volume of a sphere

It's strange that Fermat's Library stated this since the formula for volume of sphere was discovered by Archimedes more than 2000 years ago.

[1]: https://en.wikipedia.org/wiki/On_the_Sphere_and_Cylinder

Yes and no.

My understanding is that Archimedes didn't, and never could have, produced the actual formula, because that way of writing formulas didn't exist. That's discussed in the sub-thread here:

https://twitter.com/daveinstpaul/status/1213469023461138432

Additionally, the result was probably known before Archimedes, but again, writing that formula in that form probably didn't happen until Euler.

That tweet is merely about the way the formula was expressed, not about its mathematical content:

> Archimedes discovered the formula many centuries before Euler, although he could not express it in modern algebraic notation.

If Archimedes was able to express the formula in natural language (e.g. "take the radius of the hypersphere, multiply it by itself as many times as there are dimensions, ..." etc.) in a way that was unambiguously understood by other mathematicians (or natural philosophers) of the time, then that formula has definitely been discovered. Writing it in modern notation with one-character symbols for addition, multiplication (typically zero symbols!) etc. is nowhere near required for it to count as being discovered.

Yeah, the subtle difference is that discovering a geometric concept and being able to express it in a compact and flexible notation are not quite the same.

It's very easy to take for granted the sheer amount of innovation and ideas that went into being able to express that formula just because they are grade school topics today. The concept and acceptance of irrational numbers, the relationship between geometry and algebra, the use of symbols in algebra and even assigning symbols for special real numbers took centuries and a steady stream of creative leaps of genius.

An excellent and detailed explanation I highly recommend can be found here: https://hsm.stackexchange.com/a/1872.

Very interesting! It's not clear to me why it's so obvious that doubling the height of a pyramid while keeping the side constant would double the volume of the pyramid. The reason it is not obvious to me is that the proportions of the pyramid would change. The author of the post seems to think this is so obvious that it doesn't bear explanation. Can anyone explain to me why this works?

Its not specific to pyramids - its true for stretching any shape in 2D and 3D or lines 1D. Stretch it to twice its former size across a single axis and you get double the length/area/volume.

It becomes obvious in 2D if you look at how the area of a rectangle is calculated (x*y) and imagine your shape is just projected/drawn on that rectangle and stretches with it.

I did consider this, and it seemed to make it feel a little clearer—but when I first posted, I didn't feel it was obvious that the fraction of the rectangle occupied by the triangle would remain constant if you stretched it. But I think the other replies prove that that is true. Thank you.

I stumbled at that point too, but satisfied myself with a slightly more general argument than the first reply.

Imagine the pyramid is made up of a bunch of small but finite-size cubical blocks. Extending a block along any single axis doubles its volume.

Extending all the blocks vertically at the same time doubles the height and volume of the pyramid, because the volume is just the sum of the volume of all the blocks.

Mentally pass to the limit of small blocks, and you're there, or at least, I convinced myself I was there.

This feels like a perfectly intuitive 3d proof to me. Thank you.

Imagine that the pyramid is made up of a very large number of very thin slices. Now take two copies of each slice ... they are thin enough that you really haven't changed the shape, except to double the height. You've doubled the height, and doubled the volume.

It's a little like performing a riffle-shuffle on two identical packs of cards. If they aren't simply cuboids, but are identically interestingly shaped, riffling them together results in something that's the same shape, but twice the height.

Does that help?

Took me a moment. Interesting! So you're saying: Divide the pyramid into very thin slices, parallel to the base (horizontal). Above each slice, insert a duplicate of that slice. If using infinitely small slices, you have effectively doubled the height of the pyramid, and obviously you have doubled its volume.

Yes, that makes sense, and I think it's very clever! Thank you.

Okay, maybe I can accept this readily. Easy enough to prove that doubling the height of a convex 2-d shape will double the area if you stretch all of its vertical proportions to double. If that shape is on a graph, the x axis is what it normally is, and the internal height of the shape for a given value of x is given by f(x), when you double the height it will be given by 2*f(x). The integral of a function times a constant is that constant times the integral of the function, so the area inside the shape would double.

I don't remember multivariable at all, would someone be willing to offer a 3-d proof?

Makes me think of the steradian. This is used in photometery for basic things like the lumen (1 lumen = 1 candela per steradian), where there's the concept of a light source emitting in all directions uniformly, and also what amounts of that arrive at what distances onto a surface, and what amounts are absorbed or reflected, and so on.

https://en.wikipedia.org/wiki/Steradian

https://en.wikipedia.org/wiki/Lumen_(unit)

Hammer and nails, but to me the intuitive derivation of the volume formula would be through integration. Approximate slice of sphere as a cylinder, and sum up the volumes of those cylinders as their thickness approaches zero

This was all done before calculus had been developed, and the ideas here are clear precursors of the ideas later made precise by Newton and Leibniz.

The intuitive derivation is by thin slices that are thin enough to see why it's true. The trained and sophisticated version is "Well, it's just calculus, isn't it."

Wow, I might finally grok this. At h=0, the area of the dome's slice = the cylinder's slice (πRR). As h increases, the dome slice gets smaller, but by exactly πhh. I picture that smaller circle in the plane of my screen, perp to the dome slice. Somehow, for any given h, that πhh makes up for the lost area at the edge.

The bit about the Hat Box theorem still just makes me blink.

I like his proof of a the volume of a cone. I usually considered it via Euclid XII, prop 7, which was always hard to visualize. Starting with 6 pyramids in a cube is ingenious.

https://mathcs.clarku.edu/~djoyce/java/elements/bookXII/prop...

The volume and surface area of a ball are about 1/2 that of the box it came in.

This comes in handy for back-of-the-envelope calculations.

Source: http://www.vendian.org/envelope/dir0/ballbox.html

The simplest way to get the volume of a sphere or any other solid 3d object under a certain size is to fill a bucket of water to the brim, put the object in until it is completely submerged, take it out, then measure how much water is now missing. Obviously only works with smaller objects.

Experimentation can be an important tool in gathering data & error estimations to build and test a hypothesis. Reasoning is another way. There's no One True Way™ to solve a 3D geometry problem if the resulting hypothesis is demonstrably correct.

But you’d never get an exact measurement that way, since pi is irrational.

The precision of the instruments is a much bigger obstacle than the representation of the value. You would be just as uncertain that the unit cube has volume 1.000000...

4/3pir^3

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