## Is the true for any scalar function?

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

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?

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")

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
 Quote by CalcYouLater Provided that the point a lies on the closed path being integrated around?
What point are you referring to?

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 Quote by CalcYouLater 1. The problem statement, all variables and given/known data 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? 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")
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.

## Is the true for any scalar function?

 Quote by LCKurtz 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).

Thank you both for the responses.

 Quote by foxjwill 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.

 Quote by 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. $$\oint_C \phi_x\, dx = \oint_C y\, dx = -1$$ 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:

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

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

2. Relevant equations

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

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

 Quote by foxjwill 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"
 Duh, I should have been thinking about the fundamental theorem of Calculus as well as the gradient theorem. Gradient Theorem: $$\int_{a}^{b}(\nabla{f})\cdot{d{\overline{l}}}=f(b)-f(a)$$ Thanks again to foxjwill and LCKurtz for their help!