Zero curl but nonzero circulation

MHD93
Messages
93
Reaction score
0
The vector field [itex]\vec{F} = <\frac{-y}{x^2 + y^2},\frac{x}{x^2 + y^2},0>[/itex] has a zero curl, which means its circulation is zero. However

[itex]\int \vec{F}.d\vec{s}[/itex] around a unit circle on the xy plane is equal to [itex](+/-)2\pi[/itex] and not zero

Is it because F is undefined at (0,0)? No, because Stoke's theorem allows me to choose an arbitrary surface not including the origin(0, 0, 0)?
 
Last edited:
on Phys.org
Is it because F is undefined at (0,0)?
Right - more general, it is undefined at (0,0,z). As result, the domain of F is not simply connected - and the surface you want to consider does not exist.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
5K
Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 18 ·
Replies
18
Views
2K