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- Apr 13, 2013

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I am looking at the proof of this theorem:

"Let $F$ be a vector field with components $M,N$ and $P$,$F=Mi + Nj+Pk$,$M,N$ and $P$ continuous at a region $D$.Then a necessary and sufficient condition,so that the integral $\int_A^BFdR$ is independent from the road that unites $A$ and $B$ at $D$ is that there is a differentiable function $f$,so that $F=\nabla{f}=if_x+jf_y+kf_z$ at $D$.

At this case,the value of the integral is given by:

$\int_A^BFdR=f(B)-f(A)$ "

For the one direction,the proof is the following:

We suppose that there is a differentiable function $f$,such that: $F=\nabla {f}=if_x+jf_y+kf_z$.

We see that $\frac{df}{dt}=f_x\frac{dx}{dt}+f_y\frac{dy}{dt}+f_z\frac{dz}{dt} \Rightarrow \frac{df}{dt}= \nabla {f} \frac{dR}{dt}$,where $R(t)=(x(t),y(t),z(t))$.

Since, $F=\nabla {f} \Rightarrow FdR=\nabla {f} dR \frac{dt}{dt}=df \Rightarrow \int_A^BFdR=f(B)-f(A)$

How can show the other direction??