Simple line integral:confused with directions

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Aziza
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I am supposed to do the integral ∫(1/Ω)dl, where dl is the vector pointing along the loop and Ω is the distance from the origin to a point on the loop.

Let region 1 be from x=a to x=b
region 2: outer semicircle of radius b
region 3: x=-b to x=-a
region 4: inner semicircle of radius a.

Let's just look at regions 1 and 3:
in these regions, dl=dx x, x is the unit vector. dx x is a positive quantity, since it is parallel (not antiparallel) to the arrows in the above figure drawn in these regions (my path of integration). Also in regions 1 and 3, Ω=x.
Thus i set up the integral in region 1 as:

x∫(1/x)dx , with limits from a->b

and in region 3:

x∫(1/x)dx , with limits from -b->-a.

However, when I add these two results, they cancel, but the solution says they should add. I am having same problem with regions 2 and 4 (they should cancel but in my case they add). Thus I must probably be confused with signs/directions, but i do not see where my error is.

Help is much appreciated!
 
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tiny-tim said:
Hi Aziza! :smile:nooo :wink:

huh?

if you agree with my region 1 set up, then region 3 is same...except for the limits
 
tiny-tim said:
|x| :wink:

ohhh ;DDDD yes you are right, thank you!

However i am still having trouble with regions 2 and 4:

region 2 i set up as:
(1/b)∫dl2 from 0->pi

and region 4:
(1/a)∫dl4 from pi->0

Now,
dl2 = bθdθ, where θ is unit vector and positive if pointing in direction of increasing θ (ie counterclockwise)

So since in region 4, θ is pointing clockwise:
dl4 = -aθ

The sum of regions 2 and 4 however then gives me 2∫dθ from 0->pi, but it should give zero...
 
Last edited:
tiny-tim said:
that (dθ) doesn't make any sense :redface:

the RHS needs to be a vector, but dθ isn't a vector

oops you are right, that is not wat i meant. i edited the post above,
 
Aziza said:
So since in region 4, θ is pointing clockwise:
dl4 = -aθ

why the minus? :confused:

i think you're being too analytical about this whole problem …

just use common-sense: look at the diagram, and remember that ∫dl is always the length (and is always positive) :wink:
 
tiny-tim said:
why the minus? :confused:

i think you're being too analytical about this whole problem …

just use common-sense: look at the diagram, and remember that ∫dl is always the length (and is always positive) :wink:

but in the end what i am calculating is the magnetic vector potential, so the result of that integral must be a vector.

lol i have no common sense, thus i really need the formal mathematical way of doing this