# Path independence

1. Sep 19, 2014

### bmrick

my book says that the surface integral of a gradient field is path independent so long as the gradient field is continuous. This seems fishy to me. I'm envisioning a continuous gradient filed where z=grad f(x,y) and the object traced out looks like a mountain range. The equation for such a field might look like grad f = y^2 +y+5. Such a field is clearly continuous, and yet the path integral is definitely path dependant.

What does make sense to me is that if a gradient field of a conservative field exist, THEN the line integral between two points on the gradient field is path independent. And an intuitive test is to integrate the field equation you're working with and test it for conservation.
Is this right?

Last edited: Sep 19, 2014
2. Sep 21, 2014

### HallsofIvy

Staff Emeritus
Perhaps this is a question of "wording". If you know that you have a "gradient" field (not just a given vector field that might be a gradient) then, yes, as long as the field is continuous, it is path independent. I don't know what you mean by "the object traced out". Are you referring to a given path and considering that it might be non-differentiable at points? No, that would NOT be "definitely path dependent". I don't know where you got that idea. The integral is a "smoothing" process and "corners" in the path will not be important.