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.(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Path independence

Loading...

Similar Threads for Path independence |
---|

I Independent arguments of dilogarithms |

I Decomposing a Certain Exponential Integral |

**Physics Forums | Science Articles, Homework Help, Discussion**