## Proof Thata Vector Field is Conservative iff Curl of the Field is Zero

Theorem
A vector field
$\mathbf{F}$
is conservative if and only if
$\mathbf{\nabla} \times \mathbf{F} =0$
on its domain.
Proof
If
$\mathbf{F}$
is conservative then
$\mathbf{F} = \mathbf{\nabla} f$
then
$\mathbf{\nabla} \times \mathbf{F} = \mathbf{\nabla} \times (\mathbf{\nabla} f) =0$
automatically.
Conversely suppose
$\mathbf{\nabla} \times \mathbf{F} =( \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z} )\mathbf{i}- ( \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x} )\mathbf{j} + ( \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} )\mathbf{k} =0$
then
$\frac{\partial F_3}{\partial y} = \frac{\partial F_2}{\partial z} \rightarrow F_3 +a(x,z) =F_2 + b(x,y) \rightarrow F_3 a(x) =F_2 + b(x)$
(1)
$\frac{\partial F_1}{\partial z} = \frac{\partial F_3}{\partial x} \rightarrow F_1 +c(x,y) =F_3 + d(y,z) \rightarrow F_1 c(y) =F_3 + d(y)$
(2)
$\frac{\partial F_2}{\partial x} = \frac{\partial F_1}{\partial y} \rightarrow F_2 e(y,z) =F_1 + h(x,z) \rightarrow F_2 +e(z) =F_1 + h(z)$
}  (3)
(1)-(2) gives
$F_2 + b(x)=F_1 +c(y)$
(4),
(4)-(3) gives
$b(x)-e(z)=c(y)-h(z) \rightarrow b(x)=c(y)=constant$
.
Then
$a(x)=d(y)=constant$

Similarly
$e(z)=h(z)=Constant$

Hence
$f(x,y,z)$
is well defined and
$f$
exists such that
$\mathbf{\nabla} \times \mathbf{F} = 0$
automatically.