# Line integral and path dependence question

1. Oct 16, 2006

### Old Guy

Given F = iy - jx (this is my first post; not sure how you do vector notation here but I'm showing vectors in bold - hope that works). The problem is to show that this is a non-conservative force by integrating from the origin to (1,1) (ie, the path is y=x), and then do it again from the origin to (0,1) and from there to (1,1).

I just don't get how to set up the integrals - can someone help? I understand the concept behind the line integrals, but hit a brickwall trying to apply it. Thanks.

2. Oct 16, 2006

### neutrino

$$_C\int_{A}^{B}\vec{F}.d\vec{r}$$

The first part is quite simple - direct substitution. For the second part, what are x and dx =? when you go from (0,0) to (0,1)?

3. Oct 16, 2006

### Old Guy

Sorry, I don't know how to show integrals in here, but what I believe you do is that F remains the same, and you dot it with dr, which will be (dx,0) for the path from (0,0) to (0,1), and (0,dy) for the path from (0,1) to (1,1). Right so far?

4. Oct 16, 2006

### neutrino

Refer to this https://www.physicsforums.com/showthread.php?t=8997

You're right about the differentials, but what about x and y along those paths?

5. Oct 16, 2006

### Old Guy

Thanks for the reference. And I think you've hit on what I don't get - what about x and y? F is the same regardless of path, isn't it? So what's the point (and how is it calculated)? I think that's what I"m missing.

6. Oct 16, 2006

### neutrino

F = yi - xj, but what's y equal to when you're on the x-axis; in other words, when moving from (0,0) to (0,1)?

7. Oct 16, 2006

### Old Guy

y=0 along the x-axis

8. Oct 16, 2006

### neutrino

EDIT: Sorry, I've been confusing coordinates here. When you're moving from origin to (0,1) dx = 0 and x =0...it's along the y-axis. I apologise.

Last edited: Oct 16, 2006
9. Oct 16, 2006

### Old Guy

I'm not clear on what the question is now . . .

10. Oct 16, 2006

### neutrino

Okay, here's how you do it for the path from (0,0) to (0,1)

$$\int_{(0,0)}^{(0,1)}(y\hat{i} - x\hat{j}).(dx\hat{i} + dy\hat{i})$$

But since you're moving along a path in which x is a constant, in this case, zero, the above reduces to...

$$\int_{y=0}^{y=1}(ydx - xdy) = 0$$ (x = 0 and dx = 0)

Is that clear?

Last edited: Oct 16, 2006
11. Oct 16, 2006

### Old Guy

Yes, I got that.

12. Oct 16, 2006

### Old Guy

And I also don't understand how this could come to 0, because the particle was moved over a finite distance by a force - SOME amount of work should have been done!

13. Oct 16, 2006

### neutrino

Not necessarily. The work can be zero even in simple circumstances such as carrying a box ("horizontally"). There is a force, the box was moved through a finite distance, yet no (physical) work was done.

When you consider line integrals such as the above, which is a generalisation of the formula W = F.d, you evaluate the work at every point along the path, and then add it all up. It can be positive, negative, or zero.

Last edited: Oct 16, 2006
14. Oct 16, 2006

### Old Guy

Right, I do recall that. On this problem, however, the work seems to calculate to 0 for either path, which would imply the force is conservative, which my book says it is not (and I also calculated curl F to confirm this). What am I missing here?

15. Oct 16, 2006

### neutrino

Did you calculate the integral from (0,1) to (1,1)?

16. Oct 16, 2006

### Old Guy

Yes; it was also 0.

17. Oct 16, 2006

### neutrino

Can you show your work?

18. Oct 16, 2006

### Old Guy

Yes, this also came to 0.

19. Oct 16, 2006

### neutrino

The path from (0,1) to (1,1) is line y = 1, dy=0

20. Oct 16, 2006

### Old Guy

I still need a little time to figure out the LaTex stuff, but what I did was for the first step (y,-x) dot (dx,dy). dx=0, so I get -xdy, integrate to get -xy, and evaluat from y=0 to y=1 gives -x.

For the second step, (y,-x) dot (dx,dy) has dy=0, so I get ydx, integrate to get xy, and evaluate from x=0 to x=1 gives y. Summing, I get -x+y. Is the point being made that -x+y does not =0 because the values of x and y along the path do not =0? I'm looking now at my direct integration, which resulted in an integration of (xdx + ydy) from 0 to 1, for which I got 1. Is the force non-conservative because -x+y does not =1?