A false proof of drag force being conservative Consider a particle moving along some trajectory in the $x$-$y$ plane, in a viscous medium.
Then its equation of motion is given by: 
$$\mathbf{F}_d = - b \mathbf{v} .$$
it's well-known from the Gradient theorem(fundamental theorem of line integral) that if there exists a scalar-valued function $\varphi$ that satisfy:
$\mathbf{F}_d=\nabla\varphi$,then this implies $\mathbf{F}_d$ is conservative.
I wanna show through a proof by contradiction that $\varphi$ does not exist for $\mathbf{F}_d$.
Let(for the sake of Reductio ad absurdum) $\mathbf{F}_d=\nabla\varphi$.
Consider an arbitrary curve which is parameterized by the position vector $\mathbf{r}(t)=<x(t),y(t)>$.
consider that our particle is moving on that curve.
Therefore $\mathbf{F}_d$ by definition is given by:
$$\mathbf{F}_d=<-b\dfrac{dx(t)}{dt},-b\dfrac{dy(t)}{dt}>.$$
And let $\nabla\varphi$ be given by : $<\dfrac{\partial \varphi  }{\partial x},\dfrac{\partial \varphi  }{\partial y}>$
From our hypothesis $\mathbf{F}_d=\nabla\varphi$ We have that:
$$\varphi(x,y)=\int -b\dfrac{dx(t)}{dt} dx = \int -b\dfrac{dy(t)}{dt} dy. $$
Assuming that our functions are well-behaved then we get:
$$\varphi=-b( x\dfrac{dx}{dt} + y\dfrac{dy}{dt} ).$$
So Although I expected that I'd arrive at some contradiction I did not, In a sense I proved that $\mathbf{F}_d$ is conservative (Although it's not!)
So what possibly I did wrong? 
 A: The issue is that the formula that connects force and potential gets an extra term when the force depends on velocity ${\bf v}$. The formula reads (see e.g. Ref. 1)
$$\tag{1} {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}},$$
rather than just $$\tag{2} 
{\bf F}~=~ - \frac{\partial U}{\partial {\bf r}}. \qquad\qquad(\leftarrow\text{Wrong}!)$$
Velocity dependence of conservative forces and potentials is also discussed in my Phys.SE answer here. To see why the friction force
$$\tag{3} {\bf F}~=~-b {\bf v}$$
has no velocity-dependent potential $U$, see e.g. this Phys.SE post and this MO.SE post.
References:


*

*H. Goldstein, Classical Mechanics; eq. (1.58).

A: To see that your integral expression does not make any sense, imagine that $\vec{r}(t)=( x(t),y(t))$ describes a circle. Then the line integral of the force around the loop gives the change in potential energy, which should of course be zero,
$$\oint \vec{\nabla} \phi \cdot \vec{dl} = \Delta \phi =0. $$
But if you insert the actual values from your calculations,
$$\oint \vec{\nabla} \phi \cdot \vec{dl} = -b \int \vec{r}'(t) \cdot \vec{r}'(t) dt = -b\int\left| \vec{r}'(t) \right|^2 dt \neq 0. $$
