How is a line integral over any closed surface 0?

[transmini]

We just started going over line integrals in calculus, and have been told that the integral over any closed surface is 0. What I don't get is then why can we do the line integral of a circle to get 2$\pi$r? Since a circle is a closed surface, shouldn't the line integral then be 0?

 

[RUber]

You are saying to take the integral of $\int_0^{2\pi} t d\theta = 2\pi t$. However, if your integrand were a function that viewed $2\pi$ as 0, ie one which is 2pi periodic, you would also get zero.
Basically, you are comparing two different statements.
One is a loop that begins and ends at the same spot, i.e. integral from 0 to 0, the other is one that does not.

 

[pasmith]

Line integrals are taken over paths. What do you understand by a line integral over a "surface"? If you mean a line integral over the boundary of the surface, then the integral over a boundary of a closed surface vanishes because by definition a closed surface has no boundary.

A circular disc does have a boundary - its circumference.

 

[transmini]

How could we determine the difference between integrating between 2 different points that are still essentially at the same spot, such as a circle, versus a path that begins and ends at the same point that and comes out to be 0?

This site has the circle on page 2 as being 0 which is part of why I'm confused.
http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/3.-double-integrals-and-line-integrals-in-the-plane/part-b-vector-fields-and-line-integrals/session-60-fundamental-theorem-for-line-integrals/MIT18_02SC_notes_29.pdf

 

[transmini]

Line integrals are taken over paths.

I had meant to say paths instead of surfaces but couldn't think of the word for some reason.

 

[mathwonk]

line integrals (of one forms) are not always zero over closed paths, but line integrals of "exact" forms over closed paths are zero, i.e. forms of the type df. In fact a one form is exact if amnd only if the integral over every closed path is zero, if and only iof the integral from one point to another along a path is the same for every chpice of path. then one can define the function f as the integral from a fixed point to any other point, choosing any convenient path foe the integration.

a one from Pdx + Qdy is called "closed" if ∂P/∂y = ∂Q/∂x. For these forms, the integral is zero over a closed path that happens to be the boundary of a surface on which the form is defined (and smooth). In particular within a region which is say convex, or simply connected, where every closed path bounda a surface, such closed one forms are also exact. (all smooth exact one forms are closed, by the equality of mixed partials.)

 

[SteamKing]

You have to be careful with 'street-corner math'. Random people will say things mathematical, but you are never sure if they have laid all the ground work or specified all the initial conditions which make their statements true.

Always take statements which begin, 'Someone said that ... ' or 'They say that ...' as being suspect until proven otherwise.