Uni Stuff > Mathematics
volume of a hypersphere
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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:
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 .
Ilovemathsmeth:
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.
Ahmad:
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. :)
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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.
--- Quote from: Ilovemathsmeth on November 30, 2009, 12:17:20 am ---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.
--- End quote ---
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.
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