Need surface area of N-Ellipsoid.

[tohauz]

Hi guys.
What is surface area of N dimensional ellipsoid?
Any help is really appreciated.

 

[tiny-tim]

Welcome to PF!

Hi tohauz! Welcome to PF!

Hint: use a linear substitution to turn the integral into the surface area of an N-sphere

 

[tohauz]

Hi tohauz! Welcome to PF!

Hint: use a linear substitution to turn the integral into the surface area of an N-sphere

I tried. But I'm getting something which is not easily integrable

 

[tiny-tim]

Show us.

 

[tohauz]

Show us.

Suppose that ellipsoid has axis a_{1},...,a_{N}.
Then S=\int_{{\sum\frac{x^{2}_{i}}{a^{2}_{i}}<1}dS(x)-surface integral.
Then i solved for x_{N}=a_{N}(\sqrt{1-(\frac{x_{1}}{a_{1}})^{2}-...-(\frac{x_{N-1}}{a_{N-1}})^{2}}) and used that formula for evaluating the surface area, where you need evaluate N-1 dimensional volume integra (Found the partial derivatives of x_{N} w/r to x_{i} and etc). IN that integral i made a substitution , i.e. linear transformation x_{i}=y_{i}a_{i}, i=1,...N-1. I got:

Integral over {B(0,1)} of{\sqrt{1+((\frac{a_{N}}{a_{1}})^{2}-1)y^{2}_{1}+...+((\frac{a_{N}}{a_{N-1}})^{2}-1)y^{2}_{N-1}}}*Jac(Transformation), where B(0,1) is n-1 unit ball.
I think i made a mistake, but i can't find it. Thanks and by the way, what is the good textbook to brush up on these things

 

[tohauz]

Show us.

Actually, I used same idea to find the volume and I got it:
it is a_{1}*...a_{N}*meas(unit ball)

 

[tiny-tim]

Hi tohauz!

(use the X₂ tag just above the Reply box … a₁ …

and the plural of "axis" is "axes" )

Actually, I used same idea to find the volume and I got it:
it is a_{1}*...a_{N}*meas(unit ball)

That's right!

And you can do the same thing for surface area …

a₁*...a_N*surfacearea(unit ball)

 

[tohauz]

Hi tohauz!

(use the X₂ tag just above the Reply box … a₁ …

and the plural of "axis" is "axes" )


That's right!

And you can do the same thing for surface area …

a₁*...a_N*surfacearea(unit ball)

That doesn't make a sense. Because, if a_{i}=r for all i's, we don't get the surface area ball with radiuis r.

 

[tiny-tim]

That doesn't make a sense. Because, if a_{i}=r for all i's, we don't get the surface area ball with radiuis r.

oops!

should have been (a₁*...a_N)^(N-1)/N*surfacearea(unit ball)

 

[tohauz]

oops!

should have been (a₁*...a_N)^(N-1)/N*surfacearea(unit ball)

OK. Where is my mistake? Thanks
i'm having trouble with finding it

 

[Mute]

I don't think you should expect to get something easily integrable for the surface area of an N-ellipsoid. In general the surface area of just a regular 2-ellipsoid is expressible in terms of incomplete elliptic integrals. Similarly, I don't think the 'circumference' of an ellipse has a nice expression in terms of elementary functions, either. (There are some closed-form special cases)

http://en.wikipedia.org/wiki/Ellipsoid#Surface_area

 

[tohauz]

I don't think you should expect to get something easily integrable for the surface area of an N-ellipsoid. In general the surface area of just a regular 2-ellipsoid is expressible in terms of incomplete elliptic integrals. Similarly, I don't think the 'circumference' of an ellipse has a nice expression in terms of elementary functions, either. (There are some closed-form special cases)

http://en.wikipedia.org/wiki/Ellipsoid#Surface_area

I see what you are saying, but at least did I set it up correctly?

 

[tohauz]

oops!

should have been (a₁*...a_N)^(N-1)/N*surfacearea(unit ball)

Could you please tell me how you got it?

 

[tiny-tim]

Could you please tell me how you got it?

Sorry , my formula seems to be wrong …

I thought it would just be a matter of changing the coordinates, and multiplying by the appropriate factors, but Mute's link makes it clear that that doesn't work, and that the surface area, even for N = 3, is a complicated formula using "elliptic integrals".