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 similar 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.