- #1
crick
- 43
- 4
A simple method to find the potential of a conservative vector field defined on a domain ##D## is to calculate the integral
$$U(x,y,z)=\int_{\gamma} F \cdot ds$$
On a curve ##\gamma## that is made of segments parallel to the coordinate axes, that start from a chosen point ##(x_0,y_0,z_0)##.
I would like to know what are precisely the conditions that ##D## must satisfy to use this method. ##D## should be made in such way that "any point can be connected to ##(x_0,y_0,z_0)## with, indeed, a segment parallel to the coordinate axes".
But what are the sufficient mathematical conditions for $D$ in order to have this property?**
I would say that it surely has to be connected, but that seems not to be enough. For example taking
$$D= \{ (x,y) : y>x-1\} \,\,\,\, \,\,\,\,\,(x_0,y_0)=(0,0)$$
##D## is connected but I do not think that any point can be connected to ##(0,0)## via a segment parallel to the coordinate axes.
$$U(x,y,z)=\int_{\gamma} F \cdot ds$$
On a curve ##\gamma## that is made of segments parallel to the coordinate axes, that start from a chosen point ##(x_0,y_0,z_0)##.
I would like to know what are precisely the conditions that ##D## must satisfy to use this method. ##D## should be made in such way that "any point can be connected to ##(x_0,y_0,z_0)## with, indeed, a segment parallel to the coordinate axes".
But what are the sufficient mathematical conditions for $D$ in order to have this property?**
I would say that it surely has to be connected, but that seems not to be enough. For example taking
$$D= \{ (x,y) : y>x-1\} \,\,\,\, \,\,\,\,\,(x_0,y_0)=(0,0)$$
##D## is connected but I do not think that any point can be connected to ##(0,0)## via a segment parallel to the coordinate axes.