Proving a force is conservative (via path integrals)

[cyberdeathreaper]

It's always the easy questions that get me stuck....

For some reason, I'm having a mental block on how to answer this one:

Consider the force function:
F = ix + jy
Verify that it is conservative by showing that the integral,
\int F \cdot dr
is independent of the path of integration by taking two paths in which the starting point is the origin (0,0), and the endpoint is (1,1). For one path take the line x = y. For the other path take the x-axis out to the point (1,0) and then the line x = 1 up to the point (1,1).

Now I've already verified that it is conserved by taking the curl of F, but I can't seem to come to a similiar conclusion using the path integrals. Can someone help me out with this one? At the very least, if I could see the integrals themselves for each path, perhaps I could figure out where I've made my mistake. Thanks.

[ehild]

It's always the easy questions that get me stuck....

For some reason, I'm having a mental block on how to answer this one:

Consider the force function:
F = ix + jy
Verify that it is conservative by showing that the integral,
\int F \cdot dr
is independent of the path of integration by taking two paths in which the starting point is the origin (0,0), and the endpoint is (1,1). For one path take the line x = y. For the other path take the x-axis out to the point (1,0) and then the line x = 1 up to the point (1,1).


Do not forget that the work is the integral of the scalar product of force and displacement .

W = \int_{(0,0)}^{(1,1)} \vec F \cdot \vec dr = \int (F_xdx+F_ydy)=\int_{(0,0)}^{(1,1)}(xdx+ydy)

For the first path, y = x and dy = dx.
W= \int _0^1{2xdx}=1

To calculate the work along the second path, we integrate along the x axis first, than along the vertical line at x=1.

\int_{(0,0)}^{(1,1)}(xdx+ydy)=\int_{(0,0)}^{(1,1)} xdx +\int_{(0,0)}^{(1,1)}ydy=0.5*[x^2]_0^1+0.5*[y^2]_0^1=1

ehild

[cyberdeathreaper]

Thank's so much.... I see what I was doing wrong now.

Specifically, instead of
\int (F_xdx+F_ydy)
I had been using
\int ((F_x + F_y) dx dy)

Thanks again. :smile: