How can Line integrals curves not matter?

  • Thread starter Noone1982
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  • #1
Noone1982
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Thanks on the help on the other thread.

I, however, have yet another question. In the line integrals, how is it that we're integrating the various components to the limits of the curves, it seems like the curves really don't matter, just their limits.

Can someone explain how the curves are important too and not just their limits, as many functions can have the same limits ...
 

Answers and Replies

  • #2
hypermorphism
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This depends on whether you're integrating over a conservative field. If you are, the integral is path independent, but if you're not, then different paths between the same two points yield different integrals.
 
  • #3
Noone1982
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What exactly is a conservative field? How do the limits change from one type of a field to another?

How would it play into

? Ax dx + Ay dy + Az dz
 
  • #4
amcavoy
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Noone1982 said:
What exactly is a conservative field? How do the limits change from one type of a field to another?

How would it play into

? Ax dx + Ay dy + Az dz

A conservative vector field is one such that:

[tex]\oint_{C}\vec{\mathbf{F}}\cdot d\vec{\mathbf{r}}=0[/tex]
 
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  • #5
robphy
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apmcavoy said:
A conservative vector field is one such that:

[tex]\int_{C}\vec{\mathbf{F}}\cdot d\vec{\mathbf{r}}=0[/tex]

Thus rather than following the curve C, you can follow any path.

That should probably read: [tex]\oint_{\gamma}\vec{\mathbf{F}}\cdot d\vec{\mathbf{r}}=0[/tex], where [tex]\gamma[/tex] is any closed curve.
Then, [tex]\int_{C}\vec{\mathbf{F}}\cdot d\vec{\mathbf{r}}[/tex] depends only on the endpoints of C. (That is, any path with the same endpoints as C will give the same value for this integral.)

The first condition can be shown to be equivalent to [tex]\vec\nabla\times \vec F=\vec 0[/tex].
 
  • #6
HallsofIvy
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Actually, to me the surprising thing is that for some (actually MOST) line integrals, the integral does depend on the curve!

After all, in beginning calculus, we learn that, to find [itex]\int_a^b f(x)dx[/itex], we find an anti-derivative F, such that dF= f(x)dx and evaluate at the endpoints: F(b)- F(a).

Okay, if we have [itex]\int_C f(x,y)dx+ g(x,y)dy[/itex], why not just find F(x,y) such that dF= f(x,y)dx+ g(x,y)dy and evaluate at the endpoints??

The answer, of course, is that not every (in fact few) "f(x,y)dx+ g(x,y)dy" is an "exact" differential- there may not be such an F(x,y).

But for some such an F(x,y) does exist and it's just a matter of evaluating at the end points- what happens on the curve between the endpoints, or exactly what the curve itself is, doesn't matter.
 

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