Calculating Hypervolume/Hypersurface of Unit Ball in n Dimensions

In summary: It's just a rearrangement of the integral.In summary, the hypervolume/hypersurface of the unit ball in n dimensions is found by multiplying together "n" integrals of Poisson/Gauss. The first integral can be computed and then the V_n can be found using the Jacobian.
  • #1
Jerbearrrrrr
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How do you derive the hypervolume/hypersurface of the unit ball in n dimensions? I thought it'd be trivial but oh well.
I'm guessing switch to n-D spherical coordinates?
Is there an argument which doesn't involve explicitly calculating too many things? :x
 
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  • #2
Howdy, well, this argument I saw in a stat mech text: Consider 2 different ways of computing the integral (<n> arguments)

[tex] \int_{- \infty}^{+ \infty} e^{- x^2 - y^2 - z^2 - ...} dx{}dy{}dz{}... [/tex]

First it's a product of "n" integrals of Poisson/Gauss. The result of

[tex] \int_{- \infty}^{+ \infty} e^{- x^2 } dx[/tex] we know and just raise it to the power "n".

Now use the (hyper) spherical coordinates for the first integral. The result should be

[tex] \int_{0}^{+ \infty} e^{-r^2} {} \ r^{n-1} {} \ dr \times S_{n} [/tex]

S_{n} is the area of the unit ball in "n" dimensions. The first integral you can compute and then you can get S_{n}. But S_{n} it's easily linked to the V_{n} and thus you can find V_{n}.
 
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  • #3
Yeah, but um.
I "know" it's going to work out all nice, but how do you construct spherical coordinates in n dimensions, and show that the Jacobian is what it is?

I guess dV=dr.dS is reasonable. Analysis of it seems a bit more annoying though :x
 
  • #4
Jerbearrrrrr said:
Yeah, but um.
I "know" it's going to work out all nice, but how do you construct spherical coordinates in n dimensions, and show that the Jacobian is what it is?
Why bother with all that mess when you already know how the measure of a shell of constant radius scales with r?



The way I've heard it, the only tricky part is showing that the integral really can be rearranged like that without changing the value.
 
  • #5
Oops. I was kind of thinking "How do we know that's S_n"?
But that's the definition of S_n.
 

1. How do you calculate the hypervolume of a unit ball in n dimensions?

The hypervolume of a unit ball in n dimensions can be calculated using the formula: V = π^(n/2) / Γ(n/2 + 1), where n is the number of dimensions and Γ is the gamma function.

2. What is the difference between hypervolume and hypersurface?

Hypervolume refers to the volume of a shape in n-dimensional space, while hypersurface refers to the surface area of that shape. In other words, hypervolume is a measure of the space enclosed by a shape, while hypersurface is a measure of the boundary of that shape.

3. Can the hypervolume/hypersurface of a unit ball in n dimensions be visualized?

It can be difficult to visualize the hypervolume/hypersurface of a unit ball in dimensions higher than three. However, some mathematical software programs, such as Mathematica or MATLAB, have tools that can help visualize these concepts.

4. How does the hypervolume/hypersurface of a unit ball change as n increases?

As n increases, the hypervolume of a unit ball also increases, but at a decreasing rate. This means that the hypervolume continues to grow, but at a slower and slower pace for each additional dimension. The hypersurface, on the other hand, increases at a faster rate than the hypervolume as n increases.

5. Are there any real-world applications for calculating the hypervolume/hypersurface of a unit ball in n dimensions?

Yes, there are several applications for these calculations, including in the fields of physics, biology, and economics. For example, the hypervolume of a unit ball can be used to model the space occupied by particles in a gas, or the potential habitat of a species in ecology. The hypersurface can be used to calculate the surface area of a cell in biology, or the surface area of a trade zone in economics.

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