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Volumes in the 4th spatial dimension 
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#1
Feb2105, 10:51 PM

P: 82

How would you calculate the volume of a 4dimensional object? Like a hypercube, hypersphere, etc...



#2
Feb2205, 01:00 AM

P: 181

Hypercube with a side n: n^4 I guess.



#3
Feb2205, 01:07 AM

P: 112

Here check this out...
http://mathworld.wolfram.com/FourDi...lGeometry.html Remeber to book mark this website for it is very handy. 


#4
Feb2205, 02:24 AM

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P: 2,002

Volumes in the 4th spatial dimension
You can find the volume of an Ndimensional sphere of radius R by the following integral:
[tex]V_N(R)=\int\theta(R^2x^2)d^Nx[/tex] where [itex]x^2=\sum x_n^2[/itex] and [itex]\theta[/itex] is the unit step function. 


#5
Feb2205, 08:48 AM

P: 25

To find the "volume" of a 4d object you got to integrate over a 3 dimensional surface.
Take for example, if you wanted to calculate the area of a 2d object such as a circle with radius R. Then you would sum an infinite number (ie.integrate) of 1d objects, in this case the circumference of the circle between 0 and R. Can you see that the area of a circle is the infinite sum of smaller and smaller circles, until you reach the center of the circle? This is why when you differentiate the area of a circle Pi*R^2, wrt. R you get 2*Pi*R, which is the circumference, and when you integrate the circumference you get the area. Now how could you calculate the surface area of a sphere? If you get a basket ball or something you can see that the surface area of a sphere is the infinite sum of circles which starting from one pole of the surface of the sphere, get bigger, until one reaches the equator then shrink back to zero radius at the other pole. Assuming your sphere has radius 1, you'll find the circumference of your circle r units away from a pole is 2*Pi*sin(r). Integrate that between 0 and Pi and you'll get 4*Pi, which is the surface area of your sphere. Since the surface area of a sphere of radius R has units R^2, then the Surface area of a general sphere of radius R is 4*Pi*R^2. Integrating the surface area of the sphere wrt. R, you'll get the volume of the sphere, which is 4/3 Pi*R^3. The reason for this is the same as pointed out in the first example. Because taking the volume of a sphere, then subtracting the volume of a infinitely slightly smaller sphere your left with the surface area of the sphere. Anyway since you have the surface area of the sphere, if you want to calculate the volume of a hyper sphere, for all the reasons in the other examples, just integrate 4*Pi*sin(r)^2, between 0 and Pi, and you'll have the surface volume of your hypersphere. Add a R^3 term to that, and integrate wrt. R and you'll get the 4d volume of your hypersphere. If you want to find the volume of a 5d sphere, take the surface volume of your 4d sphere with radius r. You'll have a term like A*r^3. Integrate A*sin(r)^3, between 0 and Pi, your answer may be B, so the surface volumeof your 5d sphere is B*r^4. Integrate that between 0 and R and you got the volume of your 5d sphere. etc etc. You can find your self quite a nice recursive formula in such a way, which enables you to calculate the volume and surface volume of any dimensional sphere. 


#6
Feb2205, 10:11 AM

P: 84

how about hyperspherical packing like sphere packing? How would you do that?



#7
Feb2205, 10:56 AM

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P: 11,893

Daniel. P.S.Of course,it's natural to choose the system of coordinates withe the origin of axis in the center of the ball. 


#9
Feb2205, 11:50 AM

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P: 11,893

It's not mathematics the "thing" you're trying to do by ignoring the WIDELY ACCEPTED definitions of current mathematics...I don't know what it is,i'm assuming it is bulls***.
Daniel. 


#10
Feb2205, 05:04 PM

P: 347




#11
Mar305, 01:31 AM

P: 82

And what do you mean by "integrate between 0 and pi)? Thanks 


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