## Necessary and Sufficient Conditions for an Exact Differential

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
$F_1 dx + F_2 dy +F_3 dz = d \phi$