Call Us 07766496223
The exact ifferential of a function  
\[\phi\]
  is  
\[d \phi = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz\]
.
Theorem
If  
\[\mathbf{F} = F_1 \mathbf{i} + F_2 \mathbf{j} + F_2 \mathbf{k}\]
  then a necessary and sufficient condition for  
\[F_1 dx + F_2 dy +F_3 dz\]
  to be an exact differential is that  
\[\mathbf{\nabla} \times \mathbf{F}=0\]
.
Proof
Assume That  
\[F_1 dx+F_2 dy+F_+3 dz\]
  is an exat differential so that  
\[F_1 dx+F_2 dy+F_+3 dz =d \phi = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz\]
.
Then  
\[F_1 = \frac{\partial \phi}{\partial x} , F_2 = \frac{\partial \phi}{\partial y} , F_3 = \frac{\partial \phi}{\partial z}\]
  and  
\[\mathbf{F}= \mathbf{\nabla} \phi\]
.
It follows that  
\[\mathbf{\nabla} \times \mathbf{F} = \mathbf{\nabla} \times (\mathbf{\nabla} \phi )= \mathbf{0}\]

If  
\[\mathbf{\nabla} \times \mathbf{F}=\mathbf{0}\]
  then there exists a function  
\[\phi\]
  satisfying  
\[\mathbf{F}= \mathbf{\nabla} \phi\]

Then  
\[\mathbf{F} \cdot d \mathbf{r} = (\mathbf{\nabla} \phi) \cdot d \mathbf{r} =\frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz= d \phi\]

Hence  
\[F_1 dx + F_2 dy +F_3 dz = d \phi\]