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Hi guys.

What is surface area of N dimensional ellipsoid?

Any help is really appreciated.

What is surface area of N dimensional ellipsoid?

Any help is really appreciated.

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- Thread starter tohauz
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- #1

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Hi guys.

What is surface area of N dimensional ellipsoid?

Any help is really appreciated.

What is surface area of N dimensional ellipsoid?

Any help is really appreciated.

- #2

tiny-tim

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Hi tohauz! Welcome to PF!

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

- #3

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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

- #4

tiny-tim

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Show us.

- #5

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Suppose that ellipsoid has axis a_{1},...,a_{N}.Show us.

Then S=[tex]\int_{{\sum\frac{x^{2}_{i}}{a^{2}_{i}}<1}[/tex]dS(x)-surface integral.

Then i solved for x_{N}=a_{N}([tex]\sqrt{1-(\frac{x_{1}}{a_{1}})^{2}-...-(\frac{x_{N-1}}{a_{N-1}})^{2}}[/tex]) 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[tex] {\sqrt{1+((\frac{a_{N}}{a_{1}})^{2}-1)y^{2}_{1}+...+((\frac{a_{N}}{a_{N-1}})^{2}-1)y^{2}_{N-1}}} [/tex]*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

- #6

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Show us.

Actually, I used same idea to find the volume and I got it:

it is a_{1}*...a_{N}*meas(unit ball)

- #7

tiny-tim

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(use the X

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 …

a

- #8

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Hi tohauz!

(use the X_{2}tag just above the Reply box … a_{1}…

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

That's right!

And you can do the same thing for surface area …

a_{1}*...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.

- #9

tiny-tim

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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 (a

- #10

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oops!

should have been (a_{1}*...a_{N})^{(N-1)/N}*surfacearea(unit ball)

OK. Where is my mistake? Thanks

i'm having trouble with finding it

Last edited:

- #11

Mute

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

- #12

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

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

- #13

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oops!

should have been (a_{1}*...a_{N})^{(N-1)/N}*surfacearea(unit ball)

Could you please tell me how you got it?

- #14

tiny-tim

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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

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