# Simple line integral:confused with directions

1. Nov 2, 2013

### Aziza

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!

2. Nov 2, 2013

Hi Aziza!
nooo

3. Nov 2, 2013

### Aziza

huh?

if you agree with my region 1 set up, then region 3 is same...except for the limits

4. Nov 2, 2013

|x|

5. Nov 2, 2013

### Aziza

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: Nov 2, 2013
6. Nov 2, 2013

### tiny-tim

that (dθ) doesn't make any sense

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

7. Nov 2, 2013

### Aziza

oops you are right, that is not wat i meant. i edited the post above,

8. Nov 2, 2013

### tiny-tim

why the minus?

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

9. Nov 2, 2013

### Aziza

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

10. Nov 2, 2013

### tiny-tim

sorry, i got carried away, i meant to leave out the "∫"

dl is always positive (ie if you use l instead of θ, you shouldn't get confused)