Line integral $= 0$ meaning? 
Question: 
  Compute the line integral for the vector field $F(x,y) = (x^2y,y^2x)$ and the path $$r(t) = (\cos t,\sin t),\quad t \in [0,2\pi]$$

Answer:
The answer I am getting is $0$, I am fairly certain this is the correct answer. What is the meaning behind a $0$ value for a line integral. Does that mean the line doesn't exist? Or the path is symmetric?
 A: It doesn't mean anything special. Maybe one could have detected that the value of the integral is $0$ using symmetry considerations, making the actual computation of the integral superfluous.
Things start to get interesting if the integral is $0$ for all closed curves, not just a particular one. A necessary condition for this would be that your $F=(P,Q)$ satisfies
$${\partial Q\over\partial x}-{\partial P\over\partial y}\equiv0\ .$$
Since this is not the case not much more can be said in connection with this example.
A: You can interpret the line integral being zero to have some special meaning:
In physics, line integrals are used to calculate the (physical) work used to move an object (e.g. a hockey puck) along a path in some force field (e.g. the gravitational field).
A vector field in question which is a 2d-field ($F(x,y)= (x^2y,xy^2)$) might arise in Problems where the 3rd component doesn't matter, because we are moving the object on a flat table, for instance (and the form of $F$ comes from some electric field, and the puck is charged). 
If we now move the object along a given path and the path integral is zero, then we didn't need to use any work to do it, i.e. we didn't need to work against the force field.
This is similar to the fact that lifting a hockey puck takes some effort, but it's really easy to move it on a flat, slippery surface.
