Are most vector fields conservative? I started learning line integrals and vector fields, and my question is: are the majority of "nice" functions (like polynomials, trig etc) conservative?
I ask this  because of the definition:
$\vec{f}$ is conservative $\iff$ there exists a scalar field $\vec{F}$ s.t. $\nabla \vec{F} = \vec{f}$.
When equating the components to solve for $F$, it seems that almost all $\vec{f}$ can be found as the gradient of some scalar field $\vec{F}$ (assuming that we can integrate the functions of course).  
Also, I see some sources verify that the curl of $f$ is zero, concluding that $f$ is conservative, but I know this shouldn't be true right? When is this true?
 A: Polynomials/trigonometric functions/etc. can't be conservative, only vector fields can. Vector fields involving nice functions may or may not be conservative. It definitely isn't the case that these vector fields are usually conservative (as a side note, I don't know if "most vector fields are not conservative" is true in a mathematically precise sense). If we have $v = f(x,y)dx+ g(x,y)dy$, then at the very least, in order for $v$ to be conservative, we must have $f_y=g_x$, which usually isn't the case. For example, I just randomly came up with the example $v = (2x-2y)dx+(4x^2y^2)dy$, which is not conservative. 
As for your second question. A vector field being conservative always means that its curl is zero. However, the converse does not always hold. For example, consider
$$v(x,y,z) = \frac{-y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$
You can verify that its curl is zero, but the integral around the unit circle on the $x-y$ plane will yield $\pm2\pi$. On the bright side, having zero curl does imply a vector field is conservative when the domain of the vector field is simply connected (meaning that if you take any loop on the domain, then you can shrink it to a point without leaving the domain). In the above example, the domain of $v$ is $\mathbb{R}^3$ minus the $z$ axis. If you have a loop around the $z$ axis, it cannot be shrunk to a point, and so the domain is not simply connected. 
If I recall correctly, the simple connectedness condition can actually be weakened to requiring that the abelianization of the fundamental group of the domain not have elements of infinite order. 
