Cohomology - motivating example

Concept of vector fields as gradients of functions is well established. However, sometimes taking curl of vectors also results in vector fields. These concepts lead to certain simplifications when computing line integrals.

The following illustrative example is from Tu's "Introduction to Manifolds".

Let $F(x,y)=(P(x,y),Q(x,y))$ be a vector field defined on an open set $U$ in $R^2$. Let $C$ represent a parameterized curve defined by $c(t)=(x(t),y(t))$ where $t \in [a,b]$ as it moves from point $A$ to $B$ on $U$. The total work done by a particle moving along this path is given by line integral $\int_C P(x,y)dx+Q(x,y)dy$.

If vector field is a gradient of a scalar function the line integral is easy to compute using Stoke's theorem.

$F=grad(f) = (f_x,f_y)$

where $f_x =\frac{\partial f}{\partial x}$ and
$f_y =\frac{\partial f}{\partial y}$

$\int_C f_x dx+f_y dy = \int_C df = f(B)-f(A)$.

A necessary condition for $F$ to be grad of a scalar function $f$ is

$P_y = f_{xy}=f_{yx}=Q_x$.

The question is now the following:

If $Q_x-P_y=0$, is the vector field $F=(P,Q)$ gradient of some scalar function $f$ on $U$?

By correspondence between vector fields and 1-forms in $R^2$, we have,

$F=(P,Q) <-> \omega = Pdx + Qdy$

$grad f = (f_x,f_y) <->d\omega = f_x dx + f_y dy$

$Q_x-P_y=0 <-> d\omega = (Q_x-P_y)dx \wedge dy =0$.

So the question is, if $\omega =f _x dx+f_ydy$ is closed ie. $d\omega=0$ is it exact?

Answer is sometimes Yes and sometimes No and depends on $U$.