Confusion over line integrals, Green's Theoreom, Conservative fields

[bugatti79]

Folks,

1) If we have [itex]\int F \cdot dr[/itex] that is independent of the path, does that mean that the integral will always be 0?

2) For 2 dimensional problems when we evaluate line integrals directly and use Greens Theorem for every piece wise smooth closed curves C, arent we always calculating the area of the curve regardless what the functions f(x,y) and g(x,y) are in

[itex]\int_C F \cdot dr = \int_C f(x,y) dx + g(x,y) dy[/itex]

3) What is the definition of a 'smooth' curve?

 

[MisterX]

Folks,

1) If we have [itex]\int F \cdot dr[/itex] that is independent of the path, does that mean that the integral will always be 0?

No, but the path integral of a conservative field would always be 0 if the start and end points are the same.

2) For 2 dimensional problems when we evaluate line integrals directly and use Greens Theorem for every piece wise smooth closed curves C, arent we always calculating the area of the curve regardless what the functions f(x,y) and g(x,y) are in

[itex]\int_C F \cdot dr = \int_C f(x,y) dx + g(x,y) dy[/itex]

Curves do not have area. The area enclosed by a curve C (let's call this region D) in ℝ², [itex] \int_D da[/itex], is not what you would be calculating to use Green's Theorem.

Just like, in general a single integral is not equal to the length of the integration.

[itex]\int ^{b} _{a} dx [/itex] is not in general the same as [itex] \int ^{b} _{a} f(x) dx [/itex]