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Conditions For a Vector Function to be a Complete Differential

Theorem
If  
\[\mathbf{F} = (F_1 , F_2 , F_3 )\]
  is a vector field, a necessary and sufficient condition for  
\[F_1 dx + F_2 dy + F_3 dz\]
  to be a complete differential is that  
\[curl \mathbf{F} = 0\]

Proof
Assume  
\[F_1 dx + F_2 dy + F_3 = \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz = d \phi\]
  is an exact differential then
\[F_1 = \frac{\partial \phi}{\partial x}, \: F_2 = \frac{\partial \phi}{\partial y}, \: F_3 = \frac{\partial \phi}{\partial z}\]

Therefore  
\[\mathbf{F} = \frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k} = d \phi\]

Now use the identity  
\[curl grad \phi =0\]
  with  
\[grad \phi = \mathbf{F}\]
  to obtain  
\[curl \mathbf{F} =0\]

Conversely, if  
\[curl \mathbf{F} =0\]
  there exists  
\[\phi\]
  such that  
\[\mathbf{F} = grad \phi\]

Then
\[\begin{equation} \begin{aligned}F_1 dx+ F_2 dy + F_3 dz &= \mathbf{F} \cdot d \mathbf{r} \\ &= (F_1 \mathbf{i} + F_2 \mathbf{j} + F_3 \mathbf{k}) \cdot (dx \mathbf{i} + \mathbf{j} + \mathbf{k} ) \\ &=(\frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k}) \cdot (dx \mathbf{i} + \mathbf{j} + \mathbf{k} ) \\ &= \frac{\partial \phi}{\partial x} dx + \frac{\partial \phi}{\partial y} dy + \frac{\partial \phi}{\partial z} dz \\ &= d \phi \end{aligned} \end{equation}\]
 

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