Necessary and Sufficient Conditions for an Exact Differential
\[\phi\]
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
\[F_1 dx + F_2 dy +F_3 dz = d \phi\]
