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Author Topic: volume of a hypersphere  (Read 3975 times)  Share 

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/0

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volume of a hypersphere
« on: November 29, 2009, 05:01:52 pm »
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By integrating slabs the volume of an n-sphere can be worked out to be



But is there a way to simplify the formula? atm it involves a lot of computation :/

Also, is it somehow possible to extend polar/spherical coordinates to higher dimensions?

thX :)

humph

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Re: volume of a hypersphere
« Reply #1 on: November 29, 2009, 10:39:46 pm »
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Check this out:
http://vcenotes.com/forum/index.php/topic,2051.0.html
It's my favourite proof because it all falls together somewhat unexpectedly through Fubini's theorem.
You can get the surface area of the -dimensional sphere from this formula by differentiating with respect to .
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Ilovemathsmeth

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Re: volume of a hypersphere
« Reply #2 on: November 30, 2009, 12:17:20 am »
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Where are you getting these Maths qs from? I'm interested.

Of course, knowledge is limited as I haven't done Specialist...but will do so soon.
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Ahmad

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Re: volume of a hypersphere
« Reply #3 on: November 30, 2009, 12:23:47 am »
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While I admire humph's proof I don't see it as very well motivated and it doesn't appear extremely natural to me.

Here's a proof I believe you could come up with if given enough time. The first thing to notice is that the integral expression you wrote has systematically determined terminals which look surprisingly similar to each other but differ only in the number of square terms present, and this might allow for some sort of recursion. Evaluating the integral directly doesn't look promising and appears to get messy quickly.
This leads to the observation that polar and spherical coordinates naturally work best in such a scenario, so it makes sense to consider some sort of extension of spherical coordinates, and I believe if you go through the details that this works but it is not the approach I choose to take here.

Instead I'll attempt a path which involves recursion. Why do I consider this promising? Well I tried evaluating the integral in R^3 to get an idea of what I was dealing with and when I did the innermost integral I was left with,



where the region of integration is now the disc of radius R in R^2. Both the presence of x^2 + y^2 in the integrand and the circular region of integration suggest using polar coordinates to finish off the problem which I'll leave you to check.

The region of integration reduced from the ball in R^3 to the ball (disc) in R^2. This is exactly what we want to happen in a recursion, so it's a good sign. In fact, reasoning by analogy we see that each time we evaluate an inner integral we get a resulting integral which has region of integration a hypersphere of lower dimension. If we could reduce the dimension to 2D then we could switch to polar coordinates and finish off the problem. Does the idea work in R^4? At this point what I would be most worried about is that the 2 inner integrals we'd have to evaluate would be difficult, unless I could relate it to a lower dimensional case of the problem we're trying to solve. And after a bit of thought one sees that




Verify that the integral expression underlined is in fact the area of a circle with radius by writing down the integral! And in fact this generalises so that our expression for the integral in R^n is,



Which you should verify! Now substituting we get,



which upon switching to polar coordinates and evaluating, which I'll leave you to do, gives us the simple expression . Since we have the initial values and (length of interval (-1,1) is 2, and area of unit circle is ). So for example the hypervolume of a 4D ball of radius R is . From the recursion you can work out an exact value for any given dimension.

Anyway my aim of this post is that arriving at such an expression can be motivated and is something one can be expected to do without being a genius, as there weren't any rabbit out of hat ideas.  :)
« Last Edit: November 30, 2009, 02:58:08 am by Ahmad »
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Re: volume of a hypersphere
« Reply #4 on: November 30, 2009, 01:32:18 am »
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Thanks so much ahmad and humph ;D

I think I finally understand how ahmad got the recursion formula   :D
It feels so good to have a first foray into higher dimensions!

I'm finding humph's proof a little harder to understand though.
I understand how you use fubini's dydx and dxdy to equate different expressions, but I still haven't quite grasped how . It says is the centre of the ball, but here it seems like it's being used as the radius somehow...

Also, even though it seems central to the proof and it works, I'm not sure why is used. Maybe I just need time to absorb the proof first.

Where are you getting these Maths qs from? I'm interested.

Of course, knowledge is limited as I haven't done Specialist...but will do so soon.

I've been doing a bit of a crash course on multiple integrals from
http://vcenotes.com/forum/index.php/topic,20178.0.html (thankyouthankyouthankyou Damo  :))

And remembering humph and ahmad mentioning this problem a while back, I was interested in trying it

I think multiple integrals are fun because they really challenge your spatial reasoning and force you to stretch your imagination to higher dimensions.

Ahmad

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Re: volume of a hypersphere
« Reply #5 on: November 30, 2009, 01:45:01 am »
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I found multiple integrals pretty fun too because it challenges your visualization power and requires some integration skill. There are some famous volume questions which you might like to have a go at once you become more confident. Here's one: suppose you have two infinite cylinders of radius 1 in R^3, one parallel to and centered along the x-axis and one parallel to and centered along the y-axis, they intersect in a region, find the volume of this region. What if you also had a cylinder parallel to the z-axis? :)
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humph

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Re: volume of a hypersphere
« Reply #6 on: November 30, 2009, 02:50:04 am »
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\
While I admire humph's proof I don't see it as very well motivated and it doesn't appear extremely natural to me.
...which is in part why it was only discovered a few years ago :P I find it very neat because it shows the power of Fubini's theorem, but admittedly there's no immediate reason on first glance why integrating over should be relevant to determining a closed-form expression for the volume of a ball.
On the other hand, despite seeming somewhat indirect at first, it's a much less complicated way than using recursion formulae, and you don't have to split into even and odd cases.

I'm finding humph's proof a little harder to understand though.
I understand how you use fubini's dydx and dxdy to equate different expressions, but I still haven't quite grasped how . It says is the centre of the ball, but here it seems like it's being used as the radius somehow...
When you integrate over the region with respect to first, then you are keeping fixed and integrating over the region , which is by definition . So isn't the radius of the balll, it's a point inside the ball.

In general, if you have a ball of radius centred at some point , then .
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Re: volume of a hypersphere
« Reply #7 on: November 30, 2009, 02:58:40 am »
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edit: thanks humph, i'll look at your proof again tomorrow, but im off to bed for now

I'm not sure if this is right lol, I had no spare toilet paper rolls around to test this out :P

Ok, let the cylinders be A: and B: .

The place of interest is within a unit cube cube of side 2 centred at the origin, so let V(X) denote the volume of region X inside the cube.

I had trouble finding the volume of the intersection of the cylinders, so I used another approach.



It is easily seen that

Within the boundary of the cube, looks like two crossed cylinders with bits poking out.

One way to find is



To find the bits poking out, I integrated the difference of the heights of A and B over the region





I'll give the 3 cylinder version a go later (at a more appropriate time lol), but I hope this is at least close to the answer...
« Last Edit: November 30, 2009, 03:09:57 am by /0 »

Ahmad

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Re: volume of a hypersphere
« Reply #8 on: November 30, 2009, 03:05:18 am »
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I don't remember the answer off the top of my head and it's late but hopefully I'll take a closer look at this some time tomorrow :)
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Re: volume of a hypersphere
« Reply #9 on: November 30, 2009, 09:01:39 am »
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how can you guys do maths at 4am :S

/0, that's correct, but your method doesn't generalize easily to higher dimensions (im pretty sure)

The 2d case is just standard 'find the volume of surface S over area of integration D'

In this case the surface S is one of the cylinders, perhaps x^2+y^2=1
the area of integration D is a circle, maybe x^2+z^2=1




here's a hint for another way to do the 2d case if you want, then the 3d case follows from it easily.

so you have 2 toilet paper rolls intersecting each other. put them lying flat on the ground making a cross. you want to integrate over the square region in the centre. now since this volume is an intersection, the z value (height) is sometimes determined by cylinder A and sometimes by cylinder B. So split the area of integration up into different segments, depending on which cylinder the z value comes from, and exploit symmetry.