# Is the true for any scalar function?

CalcYouLater

## Homework Statement

If $$\phi$$ depends on a single position only, $$\phi=\phi(x,y,z)$$

Can I say that:

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

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

## 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")

## Answers and Replies

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

However, I'm not sure what you mean by
CalcYouLater said:
Provided that the point a lies on the closed path being integrated around?
What point are you referring to?

Homework Helper
Gold Member

## Homework Statement

If $$\phi$$ depends on a single position only, $$\phi=\phi(x,y,z)$$

Can I say that:

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

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

## 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")

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.

$$\oint_C \phi_x\, dx = \oint_C y\, dx = -1$$

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

foxjwill
Not sure what you mean when you say φ "depends on a single position only"
I think he means that φ depends on a single parameter (e.g. only the x-coordinate).

CalcYouLater
Thank you both for the responses.

However, I'm not sure what you mean by

What point are you referring to?

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.

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.

$$\oint_C \phi_x\, dx = \oint_C y\, dx = -1$$

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

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:

## Homework Statement

Irrotational field theorem

Given that:

a.) $$\overline{\nabla}\times\overline{F}=0$$
b.) $$\oint{\overline{F}\cdot{d{\overline{l}}}=0$$

Show that a$$\rightarrow$$b

## Homework Equations

$$\overline{\nabla}\times\overline{F}=0\Leftrightarrow{\overline{F}}=\overline{\nabla}V$$

## 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 $\nabla V$ is a path-independent vector field.

CalcYouLater
Well, you could use the fact that $\nabla V$ is a path-independent vector field.

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

$$\overline{\nabla}V{\bot}{d\overline{l}}$$

For constant "V"

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

$$\int_{a}^{b}(\nabla{f})\cdot{d{\overline{l}}}=f(b)-f(a)$$