# Need surface area of N-Ellipsoid.

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

tiny-tim
Homework Helper
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 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
Homework Helper
Show us. 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

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
Homework Helper
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*...aN*surfacearea(unit ball) 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*...aN*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
Homework Helper
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*...aN)(N-1)/N*surfacearea(unit ball) oops! should have been (a1*...aN)(N-1)/N*surfacearea(unit ball) OK. Where is my mistake? Thanks
i'm having trouble with finding it

Last edited:
Mute
Homework Helper
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 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?

oops! should have been (a1*...aN)(N-1)/N*surfacearea(unit ball) Could you please tell me how you got it?

tiny-tim
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".