Surface area of sphere in n dimensions

In this sequel to the video "Volume of a ball in n dimensions", I calculate the surface area of a sphere in R^n, using a clever trick with the Gaussian function exp(-1/2 |x|^2). Along the way, we discover the coarea formula, which is the analog of polar coordinates, but in n dimensions. Finally, I show why the volume of a ball is in fact the integral of the surface area of the sphere. Enjoy!

[@drpeyam]

Note: There’s a typo at the end: there shouldn’t be a 2 on the numerator in the final answer, because n/2 Gamma(n/2) = Gamma(n/2 + 1)

[@ffggddss]

Right! I caught that. The 2 in the numerator going into that, joins with the N in the denominator, to yield (N/2)Γ(N/2) = Γ(N/2 + 1) in the denominator. *Please "PIN" this comment of yours, so that it stays at the top of the comments list. ;-)*

[@drpeyam]

ffggddss Thanks for letting me know!!! I thought I pinned it, but apparently it didn’t work! :)

[@ffggddss]

OK, it worked that time, and this should save you at least a few repetetive comments that keep pointing this out. And thanks for another nice excursion through N-balls, and (N-1)-spheres!

[@mahdieskandari8328]

Lovely work Dr. Peyam! I can watch your series 2X speed play and it still works. :)

[@furqanhabib125]

Can you make vedio for surface area or volume in n+2 dimensionl of sphere

[@ishaanthareja6369]

What an amazing professor! Kudos to you, explained really well! Thank you for this!

[@drpeyam]

Thank you!!

[@dgrandlapinblanc]

Merci. Encore une démonstration sympa. C'est très agréable de suivre tes développements.

[@TheMauror22]

I do had a lot of spherical fun!! Amazing video!!!

[@frankjohnson123]

Dr. Peyam, could you explain why around 6:30 you divide the angles by the radius to get the scaling of the hypersurface? It seems like the two should be independent because the angular bounds are the same regardless of the radius.

[@Ayrton-Da-Silva]

Very interesting video! btw when I heard you talking about using spherical coordinates in integrals of higher dimensions I began searching about it and I found about hyperspherical coordinates, then I strumbled with a question I couldnt find! In R3 the limits of the spherical coordinates of a sphere are ( 0≤r≤R , 0≤theta≤2π , 0≤phi≤π ) so I can put the limits in the integral, but what happens with the limit of the fourth coordinate so I can use it to calculate the integral of the volume of a sphere in the fourth dimension, using spherical coordinates???

[@yuvalpaz3752]

in Rn we have 0≤r≤R , 0≤phi_(n-1)≤2π , 0≤phi_k≤π where k is all the integers between 0 and n-1. look here en.wikipedia.org/wiki/N-sphere#Spherical_coordinates

[@Ayrton-Da-Silva]

ohh so its just like phi repeats itself, thx! i saw that part earlier but couldnt find the answer

[@furqanhabib125]

Dr.Peyam what is the surface area or volume in n+2 dimensional of a sphere .

[@2752RUMI]

Thank you so so much Professor!

[@justwest]

Exactly what we were just doing in calc 3, nice

[@alimghazzawi3700]

Ok surface area is just 2 dimensional object so can u find the volume surrounding the ball in n dimensions by volume i mean 3d not like last video because i think that any object has a property of each dimension bellow it like circle has its radius and its circumference and ball has a surface area and radius and circumference and i think this well carry on in any dimension n so my question is about the volume surrounding a ball in n dimension like i said volume i mean 3D . If u read it please response to me

[@alimghazzawi3700]

and correct me if i am wrong

[@yuvalpaz3752]

nice one, you have a little mistake: at the end Ngamma(N/2)=2gamma(N/2+1) which would cancel the other 2 and you will get that V=R^n*pi^(n/2)/gamma(n/2+1) i like more the recursive relation between the the volume and the surface: the unit n-ball is the union of (n − 1)-sphere shells with similar center so V_(n+1)=integral from 0 to 1 of S_n R^n dR hence V_(n+1)=S_n/(n+1) also the unit (n + 2)-sphere as a union of "donut"(tori) to get(after change of variables) S_(n+2) the integral from 0 to pi/2 S_1 * S_n * R^n cos(a)da by returning cos(a)da to the original variable and noticing that S_1 is a constant(=2pi) we get the integral from 0 to 1 S_n * R^n dR*2pi=2piV_(n+1) by setting n=n+1 we get S_(n+1)=2piV_n in the end we get the following: V_0=1 V_1=2R V_(n+1)=S_n/(n+1) S_0=2 S_1=2piR S_(n+1)=V_n*2pi by noticing that V_n=2pi*V_(n-2)/(n+2) we can use induction for even n and odd n(although it is hard to combine the solution to a general solution using gamma) (Psss... btw, i am still waiting)

[@drpeyam]

Yuval Paz Correct :) I was gonna use the recursive approach, but this approach is nice because it gives us an explicit formula!

[@lorendisney5068]

Does this work for fractional dimensions?

[@drpeyam]

The recursive formula still holds, but to find the answer you’ll need the volume of the unit ball in alpha dimensions where alpha is between 0 and 1

[@tonykarp5981]

Maybe, a brief course on complex analysis (less proofs, more of intuition) instead of linear algebra bc everyone can read on linal and understand everything but that doesn’t work with complex analysis(((( I mean you used complex analysis several times and I think few people understand that magic. Thank you for such great videos!!!

[@drpeyam]

There is no complex analysis in this video, though!

[@tonykarp5981]

Dr. Peyam's Show this video is the last one, I don’t know how KZbin notifications work, so I thought it’s more likely that you see my comment if I write here

[@drpeyam]

I see! I’ll think about it :)

[@furqanhabib125]

And can you make a vedio on surface area or volume in n+2 dimension space time of sphere

[@Zeboss321]

Thank you so much for this video Dr Peyam! Awesome as always!

[@drpeyam]

Zeboss321 Yeah, I was just testing if people are watching the video until the end ;) Thanks for noticing, I was thinking there was something fishy going on!

[@ftcbrandt]

The limit for large N is interesting.

[@MrRyanroberson1]

"the surface area of a sphere in N dimensions is the set of all points equidistant to the center in that many dimensions, which is notably always of dimension one lower than its living-space, because the volume itself is supposed to be of N dimensions and the surface area is just defined as the mathematically thin bounding region of that volume." more concise explanation? maybe it's too technical since i'm already versed in polydimensional mathematics :|

[@drpeyam]

That doesn’t explain why the dimension is exactly one lower, just that it’s lower! In theory the region can be so thin that it’s two dimensions lower!

[@MrRyanroberson1]

indeed.... i guess then an additional proof would be needed to show how a string doesn't contain a 3d ball? i guess the fastest way to convey the intuition would be that a point doesn't envelop a plate.

[@ffggddss]

Isn't it pretty easy to prove that the hypersurface (boundary) of an n-dimensional region in Εⁿ, that is in some sense, well-behaved, is necessarily an (n-1)-dimensional manifold?

[@drpeyam]

I wouldn’t say it’s that obvious, but there is the implicit function theorem which says that locally the sphere is the graph of a function, but there are very irregular sets whose boundary might be less than n-1 dimensional.

[@ffggddss]

Yes, I was thinking along the lines that if everything it takes to specify the surface is well-behaved - continuous and all that - then since the ball is an n-dim region, and its boundary takes one real parameter to constrain, by means of a continuous constraint equation, the surface can be found to be (n-1)-dim. I don't know the rigorous details of this; I just have a hunch that it works something like that. Fred