Uni Stuff > Mathematics
volume of a hypersphere
Ahmad:
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? :)
humph:
--- Quote from: Ahmad on November 30, 2009, 12:23:47 am ---\
While I admire humph's proof I don't see it as very well motivated and it doesn't appear extremely natural to me.
--- End quote ---
...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.
--- Quote from: /0 on November 30, 2009, 01:32:18 am ---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...
--- End quote ---
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 .
/0:
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...
Ahmad:
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 :)
zzdfa:
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.
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