Finding a Scalar Function on a Bounded Surface

[Eidos]

Hi guys and gals

This is a conceptual question. Let's say I have a scalar function, $$f(x,y,z)$$ defined throughout $$\mathbb{R}^3$$. Further I have some bounded surface, S embedded in $$\mathbb{R}^3$$.

How would I find the function f, defined on the surface S?

Would it be the inner product of f and S, $$<f|S>$$ or a functional composition like $$f \circ S$$?

 

[mathman]

f is a scalar, so inner product of S and f makes no sense. I don't know what you have in mind by functional composition

 

[ice109]

you mean you want parameterize f by s? as in restrict f to s? like for the purposes of a surface integral?

 

[Eidos]

f is a scalar, so inner product of S and f makes no sense.

From what I understand the inner product <f|g> is
$$\int_{-\infty}^{\infty}f(t)g^{*}(t)dt$$.

The mistake I made was to think that they are scalar functions as well even though f and g are complex functions. Sorry about that. The closest thing I've come to inner products for functions was the orthonormality of the basis functions for Fourier series.

you mean you want parameterize f by s? as in restrict f to s? like for the purposes of a surface integral?

This is exactly what I had in mind. Sorry I should have been more explicit where I was going with it.

I understand what we are doing if we have a vector field $$\textbf{F}$$ and want to find out how it permeates (eg. flux through a surface) a surface S but dotting it with the unit normal of the surface and integrating on the surface. This is actually what made me think of the inner product:
$$\iint\textbf{F}\cdot\textbf{n}\,\mathrm{dS}$$

Thanks for the replies

 

[HallsofIvy]

That is the "inner product" only if you are thinking of f and g as vectors in L₂.

You have a function, f(x,y,z), and are given a surface S. You don't say how you are "given" the surface but since it is two dimensional, it is always possible to parameterize it with two variables: on S, x= x(u,v), y= y(u,v), z= z(u,v). Replace x, y, and z in f with those: f(x(u,v),y(u,v),z(u,v).

For example, suppose you have the parabolic surface z= x²+ 2y² and some function f(x,y,z). Then you can take x and y themselves as parameters and, restricted to that surface, your function is f(x,y,x²+ 2y²).

 

[Eidos]

Thanks HallsofIvy that cleared things up :)