# Scalar Function on a Surface

1. May 1, 2008

### Eidos

Hi guys and gals

This is a conceptual question. Lets 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 $$f \circ S$$?

2. May 1, 2008

### 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

3. May 1, 2008

### ice109

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

4. May 2, 2008

### Eidos

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 aswell 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.

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

5. May 2, 2008

### 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).

6. May 2, 2008

### Eidos

Thanks HallsofIvy that cleared things up :)

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