# Interpreting path independent line integrals in terms of work done

I understand that integrating a force $\boldsymbol{F}$ along a curve $C$ represents the work done by that force. I am, however, struggling to interpret this in terms of path independent line integrals.

Say my force (conservative vector field) moves my particle from $(0,0)$ to $(1,1)$ along the curve $y=x$, for example. Say the same force moves my particle from $(0,0)$ to $(0,5)$ along $x=0$ then from $(0,5)$ to $(1,5)$ along $y=1$ then from $(1,5)$ to $(1,1)$ along $x=1$. We know that the result of the work done calculation will be the same since the vector field is conservative. This seems odd to me as the particle has 'moved' a greater distance in the second example.

Could anyone help to explain this?

I think I understand how a closed loop integral relates to a double integral over the area enclosed by the curve (Green's theorem), but I might yet have (potentially stupid) questions about this as well.

Any help would be great - thanks!

For a prominent result, we can say that any conservative field $F$ can be written as $$\vec F=\nabla \phi$$
Now if you move a particle from the point $P_1$ to $P_2$, the work done is as follows: $$W=\oint_C \vec F \cdot d\vec r$$ $$=\int_{P_1}^{P_2} \nabla \phi \cdot d\vec r$$ $$=\int_{P_1}^{P_2} \left(\frac{\partial \phi}{\partial x}\hat i+\frac{\partial \phi}{\partial y}\hat j+\frac{\partial \phi}{\partial z}\hat k\right) \cdot \left(\hat i dx+\hat j dy+\hat k dz\right)$$ $$=\int_{P_1}^{P_2} \left(\frac{\partial \phi}{\partial x}dx+\frac{\partial \phi}{\partial y}dy+\frac{\partial \phi}{\partial z}dz \right)$$ $$=\int_{P_1}^{P_2} d\phi$$ $$=\phi(P_2)-\phi(P_1)$$
So the line integral is independent of what $C$ is, it only depends on $P_1$ and $P_2$.