Let's return to our expression for a total differential of a scalar function, given above:

Then

independent of path! The integral depends only on the end points for any total differential that is integrated! Hence we know that:

This should seem very familiar to you. Suppose for a well-behaved scalar function . Then:

independent of path. In introductory mechanics you probably went from the proposition that the work integral was independent of path for a conservative force to a definition of the potential energy, but as far as vector calculus is concerned, the

Robert G. Brown 2017-07-11