# N-dimensional sphere

This was done while I was in a stat-mech class a long, long time ago.......didn't understand it very well at the time. It was an aside kind of thing the prof did, but I'm still curious about this:

How do you find the surface area of an n-dimensional sphere?

Thanks!

DaveC426913
Gold Member
Just guessing,
The surface is the derivative of the volume.

In 3D, volume is: $$4 \pi r^3$$ and surface is: $$\frac{4}{3} \pi r^2$$

So,
A 4D sphere's surface area will be : $$4 \pi r^3$$ while its volume will be $$16 \pi r^4$$

I think.

I was kind of hoping someone could explain the derivation to me. I'm not sure if the link to mathworld is the solution or not. Even if it is I don't understand that any better without some kind of explanation.

I'm 99.999% sure the other response is not going in the right direction, but I appreciate the responses!

mathwonk
Homework Helper
2020 Award
do it by inductive reasoning. i.e. to get the surface are you just need to differentiate the volume wrt radius. to get the volume you integrate the n-1 volume of a slice.

try to recall how to get the volume ofa 3 sphere by integrating area of circular slices, then apply that one dimension up.

saltydog
Homework Helper
fizixx said:
I was kind of hoping someone could explain the derivation to me. I'm not sure if the link to mathworld is the solution or not. Even if it is I don't understand that any better without some kind of explanation.

I'm 99.999% sure the other response is not going in the right direction, but I appreciate the responses!

I think the following web site is a good one:

Area and Volume formulas for n-dim spheres and balls

I'll go over the integrals too!

Edit: Ok I did:

Hey guys, I found the volume formulas for n-balls interesting: simply integrate a cross-section of the object over appropriate bounds. For the ball, it was:

$$V_2=2\int_0^{\pi/2} (line)dheight$$

$$V_3=2\int_0^{\pi/2} (circle)dheight$$

$$V_4=2\int_0^{\pi/2} (sphere)dheight$$

and so on if you check the reference I gave above. I wonder if this is the case in general. That is, if I calculate the volume of my coffee cup then would the volume of a 4-D coffee cup be:

$$V_4=\int_a^b \text{(3-D coffee cup)} dp$$

for appropriate values of a, b, and p.

(Hope I don't make it confussing for you Fizixx).

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saltydog
Homework Helper
Got some questions:

Can we say in general to calculate the volume of ANY object in n-dimensional space, integrate the n-1 dimensional object over appropriate bounds? What then is the volume of the n-D torus? Same dif? I would say so. Maybe we need to calculate it.

Also, I found it interesting in the MathWorld reference that the volume of the sphere reaches a maximum value at around 5-D and then drops getting closer and closer to zero as the dimension is increased. Is that the case with all n-D volumes? Does this mean that my coffee cup holds smaller and smaller amounts of n-D coffee?

I can see it now: Salty's n-D Coffee Cafe' . . . where the coffee never runs out.

saltydog...and others.....

Great in concept, but it does not answer my question, unfortunately. I think I asked my question on the wrong board.

Providing an aswer that tritely says to just integrate your way up the dimension-ladder despite isn't sufficient, but I do apreciate the responses a great deal, and thanks to everyone for sending a response, but I'm closing this thread and going elsewhere for further inquiry.

Good day all, and hapy thoughts to everyone!

rachmaninoff
fizixx said:
Great in concept, but it does not answer my question, unfortunately. I think I asked my question on the wrong board.

Providing an aswer that tritely says to just integrate your way up the dimension-ladder despite isn't sufficient,

It gives an explicit formula for the hypercontent ('volume') of the n-ball, in terms of the gamma function:

$$V_n(r) = \frac{\pi ^{n/2} r^n}{\Gamma \left( 1 + \frac{1}{2}n \right)}$$

From which the 'area' of a hypersphere's surface naturally follows, in the usual way:

$$S_n(r) = \frac{ dV_n }{ dr } = \frac{n \pi ^{n/2} r^{n-1}}{\Gamma \left( 1 + \frac{1}{2}n \right)}= \frac{2 \pi ^{n/2} r^{n-1}}{\Gamma \left( \frac{1}{2}n \right)}$$

If you don't like this, there's a table at the bottom of the page which works out the first ten dimensions; and a more complete listing is Sloane's A072478 and A072479 .

edit: More of the same is here (also from Mathworld).

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rachmaninoff