Is the true for any scalar function?

CalcYouLater

1. The problem statement, all variables and given/known data

If [tex]\phi[/tex] depends on a single position only, [tex]\phi=\phi(x,y,z)[/tex]

Can I say that:

[tex]\oint{\frac{\partial\phi}{\partial{x}}dx=\oint{d\phi}=[\phi]_{a}^{a}=0[/tex]

Provided that the point a lies on the closed path being integrated around?

2. Relevant equations



3. The attempt at a solution

I am 99.99% sure I can. I am in the middle of "showing" how one theorem implies another, and I wasn't sure if I could knock the partial dx off with the total dx in any case under the conditions above. Thanks for any help. (Sorry, I can't seem to get my latex correct. The second to last term is intended to be phi evaluated from a to a. Also the title should read "Is this true for any scalar function")

foxjwill

Well, since [itex]\phi[/itex] depends only on [itex]x[/itex],
[tex]
\frac{\partial\phi}{\partial y} = 0 = \frac{\partial\phi}{\partial z}.
[/tex]
So,
[tex]d\phi = \frac{\partial\phi}{\partial x} dx + \frac{\partial\phi}{\partial y} dy + \frac{\partial\phi}{\partial z} dz = \frac{\partial\phi}{\partial y} dx.[/tex]
Hence,
[tex] \oint d\phi = \oint \frac{\partial\phi}{\partial x}dx = \int_a ^a\frac{\partial\phi}{\partial x}dx = 0.[/tex]

However, I'm not sure what you mean by
What point are you referring to?

LCKurtz

Not sure what you mean when you say φ "depends on a single position only" but I think the answer is no. Consider φ = xy and C the boundary of the unit square.

[tex]\oint_C \phi_x\, dx = \oint_C y\, dx = -1[/tex]

because only the top side of the square contributes a nonzero value.

foxjwill

I think he means that φ depends on a single parameter (e.g. only the x-coordinate).

CalcYouLater

Thank you both for the responses.

Sorry, I should have been more clear. I intended to say that "a" is a point on the path. That was bad use of wording on my part.

When I said that φ depends on a single position only, I meant to imply that it was something like a potential, or temperature distribution. I can see how the way I worded it is confusing.

After thinking about LCKurtz's scenario with the unit square I realize that I have totally butchered this question.

Here is what I should have done from the start:

1. The problem statement, all variables and given/known data

Irrotational field theorem

Given that:

a.) [tex]\overline{\nabla}\times\overline{F}=0[/tex]
b.) [tex]\oint{\overline{F}\cdot{d{\overline{l}}}=0[/tex]

Show that a[tex]\rightarrow[/tex]b


2. Relevant equations

[tex]\overline{\nabla}\times\overline{F}=0\Leftrightarrow{\overline{F}}=\ove rline{\nabla}V[/tex]





3. The attempt at a solution

I know that Stoke's theorem will show this in an instant. I wanted to try and show it using the equations listed under relevant equations. My thought was that by writing the vector indicated by "F" as the gradient of some scalar, I could show that integrating along a closed path would give me a null result for any scalar chosen.

foxjwill

Well, you could use the fact that [itex]\nabla V[/itex] is a path-independent vector field.

CalcYouLater

Hmm, does that mean it is as simple as saying:

[tex]\overline{\nabla}V{\bot}{d\overline{l}}[/tex]

For constant "V"

CalcYouLater

Duh, I should have been thinking about the fundamental theorem of Calculus as well as the gradient theorem.

Gradient Theorem:

[tex]\int_{a}^{b}(\nabla{f})\cdot{d{\overline{l}}}=f(b)-f(a)[/tex]

Thanks again to foxjwill and LCKurtz for their help!