Surface Area of an N-Dimensional Sphere

[fizixx]

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]

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.

 

[rachmaninoff]

Not quite.

Read this article: http://mathworld.wolfram.com/Ball.html

 

[fizixx]

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]

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]

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 website 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).

 

[saltydog]

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.

 

[fizixx]

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]

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,

Did you read my link? ( http://mathworld.wolfram.com/Ball.html )


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).

 

[rachmaninoff]

By the way, you can substitute the Gamma function with an ordinary factorial (for the usual, integer dimensions), this result is also given in that article:

http://mathworld.wolfram.com/Hypersphere.html

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