# Volumes in the 4th spatial dimension

by Aki
Tags: dimension, spatial, volumes
 P: 82 How would you calculate the volume of a 4-dimensional object? Like a hypercube, hypersphere, etc...
 P: 181 Hypercube with a side n: n^4 I guess.
 P: 112 Here check this out... http://mathworld.wolfram.com/Four-Di...lGeometry.html Remeber to book mark this website for it is very handy.
HW Helper
P: 2,004

## Volumes in the 4th spatial dimension

You can find the volume of an N-dimensional sphere of radius R by the following integral:
$$V_N(R)=\int\theta(R^2-x^2)d^Nx$$
where $x^2=\sum x_n^2$ and $\theta$ is the unit step function.
 P: 83 how about hyperspherical packing like sphere packing? How would you do that?
HW Helper
P: 11,718
 Quote by Galileo You can find the volume of an N-dimensional sphere of radius R by the following integral: $$V_N(R)=\int\theta(R^2-x^2)d^Nx$$ where $x^2=\sum x_n^2$ and $\theta$ is the unit step function.
The volume of any sphere (any # of dimensions) is ZERO...You were probably referring to a ball... An N-1 dimensional ball... (Assuming it is open,the equation would be $\sum_{i=1}^{n} x_{i}^{2} < R^{2}$ )

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.
HW Helper
P: 2,004
 Quote by dextercioby The volume of any sphere (any # of dimensions) is ZERO...
Here we go again...
 HW Helper Sci Advisor P: 11,718 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.
P: 347
 Quote by Galileo Here we go again...
My sentiments exactly.
P: 82
 Quote by damoclark 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.
I'm already lost here, lol. How did you get 4*pi when you integrate 2*pi*sin(r)? Shouldn't it be 2*pi*(-cos r) if you take the antiderivative?
And what do you mean by "integrate between 0 and pi)?

Thanks

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