Proof That Any Vector Field Represented by a Symmetric Jacobian Matrix is Conservative

Any vector field  
  represented by a symmetric Jacobian matrix on a region  
  is a conservative field.
Take any two points  
\[x_0 , \: x_1\]
  and construct a rectangle with edges parallel to the coordinate axis in 
. The  
  be the set of all paths from  

Any element of  
  consists of a path which we may represent as by the directions and lengths of the path segment at each vertex on the path eg  
\[ x_{a_1}e_{a_1} x_{a_2}e_{a_2} ... x_{a_m}e_{a_m}\]
  being the alth basis vector for the ith axis. Take any two paths
\[\gamma_p : x_{p_1}e_{p_1} x_{p_2}e_{p_2} ... x_{p_m}e_{p_m}\]

\[\gamma_q : x_{q_1}e_{q_1} x_{q_2}e_{q_2} ... x_{q_m}e_{q_m}\]

Each path segment occur once in each of the two paths since each is the side of a triangle, opposite sides being the same length, so each path can be transformed into the other by a sequence of interchanging of adjacent vectors. Consider the closed path
  consisiting of the segments  
\[x_{p_s}e_{p_s} x_{p_t}e_{p_t}\]
  formed into a reacngle.
\[\oint_C \mathbf{F} d \mathbf{x} = \oint_C F_p dx_p + F_q dx_q\]

Green's Theorem tells us
\[\oint_C \mathbf{F} d \mathbf{x} = \int_S ( \frac{\partial F_q}{\partial x_p} - \frac{\partial F_p}{\partial x_q}) dx_p \: x_q =0\]

Since the Jacobian matrix is symmetric, so  
\[\frac{\partial F_q}{\partial x_p} = \frac{\partial F_p}{\partial x_q}\]
This mean a change of path by swapping any two adjacent path segements does not change the integral. Since pthe paths can be changed into each other by a sequence of such operation, each integral gives the same result and the field is conservative.