# Vector calc proof question

1. Sep 22, 2008

### jackiefrost

This isn't homework. I was browsing through my old James Stewart multivariable calc textbook and am having a mental block concering an aspect of the "proof", shown in the file attachment, below. I've highlighted the portion giving me trouble.

My confusion concerns how the first integral (the line integral for C1) can be regarded as independent of x. The integral isn't independent of r which I assume would be dependent on both x and y.

http://home.comcast.net/~ut1880h/Files/stewart_theorem.jpg

jf

Last edited: Sep 22, 2008
2. Sep 22, 2008

### tiny-tim

Hi jackiefrost!

The integral is $$\int_{(a.b)}^{(x_1,y)}\bold{F}\cdot d\bold{r}$$

This is a function of x1 and y only.

If you increase or decrease x a little (keeping y the same), you can still use the same x1 … so, locally, x1 is independent of x (and so is y, of course).

3. Sep 23, 2008

### jackiefrost

Hi tt,
Thanks for the insight. Once was I blind but now I see...

I'm starting to appreciate how cleverly this proof is structured to efficiently arrive at the desire end. It kinda tickles my head in a way that feels good (if ya know what I mean)

jf

4. Sep 23, 2008

### tiny-tim

thou once wast lost, but now art found …

Hi jf,

it is a champagne amongst proofs!

hallelujah!

5. Sep 23, 2008

### mathwonk

this is a trivial fact that follows instantly from the FTC. write your field as Pdx + Qdy, and then define the primitive function f(p) as the path integral of Pdx + Qdy taken along any path from a fixed point a to p.

all you need to do is show the x partial of f is P and the y partial is Q.

since the integral is the same for all paths, you choose the path to make this calculation easy.

e.g. to calculate ∂f/∂x make the path piecewise rectangular, with the last bit parallel to the x axis.

then the part of the integral for Qdy is zero (since y is not changing along a path parallel to the x axis), and the derivative of the Pdx bit is P by the FTC. etc...