How to prove that a vector field with holes in its domain is conservative without finding a potential? I have to prove that $$\vec{F}=\frac{\vec{r}}{r^2}$$ is a conservative vector field without finding it's potential function.
However, $\vec{F}$ has a hole in its domain, at $(x,y,z)=(0,0,0)$, so I can't just check it's curl.
How do I go about proving this? 
 A: Assuming that $r$ refers as usual to the radial spherical coordinate, for this example it still suffices to show that $\operatorname{curl} \vec{\bf F} = 0$. It's true that the domain of $\vec {\bf F}$ has a hole, but the domain is still simply connected: Any loop in the domain, say, based at a point $\vec{\textbf{x}}_0$ can be continuously deformed to the constant path at $\vec{\textbf{x}}_0$.
Alternatively, in spherical coordinates $(r, \theta, \varphi)$, the gradient is $$\operatorname{grad} f = f_r \vec{\bf r} + \frac{1}{r} f_{\theta} \vec{\bf \unicode[Arial]{x3B8}} + \frac{1}{r \sin 
\theta} f_{\varphi} \vec{\bf \unicode[Arial]{x3C6}}.$$ In particular, the gradient of a radial function $f(r)$ is $$\operatorname{grad} (f(r)) = f_r(r) \vec{\bf r} ,$$ so the Fundamental Theorem of Calculus guarantees that any continuous radial vector field (not defined at the origin, where spherical coordinates behave badly) is conservative.

For a vector field $\vec{\bf G}$ whose domain is not simply connected, $\operatorname{curl} \vec{\bf G} = 0$ is inconclusive. For example, both $$\vec{\bf R} := \frac{x}{x^2 + y^2} \vec{\bf x} + \frac{y}{x^2 + y^2} {\vec{\bf y}}$$ and $$\vec{\bf V} := \frac{-y}{x^2 + y^2} \vec{\bf x} + \frac{x}{x^2 + y^2} {\vec{\bf y}}$$ have domain $\{(x, y, z) : (x, y) \neq (0, 0)\}$ and have zero curl, but $\vec{\bf R}$ is conservative and $\vec{\bf V}$ is not. (In cylindrical coordinates $(R, \theta, z)$, $\vec{\bf G} = \frac{1}{R} \vec{\bf R}$ and $\vec{\bf V} = \vec{\bf \unicode[Arial]{x3B8}}$.)
In this particular case, the domain has a single hole that prevents it from being simply connected (more precisely, its fundamental group is a free with a single generator), so to check whether the vector fields are conservative it suffices to check whether the integral along a single suitable curve $\gamma$ vanishes. Here, suitable just means that any loop $\alpha$ is continuously deformable to the curve $n \gamma$ for some $n$, where $n \gamma$ denotes tracing out the loop $n$ times (or $-n$ times backward if $n < 0$); informally, $n$ just counts the (signed) number of times $\alpha$ wraps around the hole. Then the vanishing of the curl guarantees that
$$\int_{\alpha} \vec{\bf G} \cdot d\vec{\bf s} = \int_{n \gamma} \vec{\bf G} \cdot d\vec{\bf s} = n \int_{\gamma} \vec{\bf G} \cdot d\vec{\bf s} .$$
So, $\vec{\bf G}$ is conservative, that is, $\int_{\alpha} \vec{\bf G} \cdot d\vec{\bf s}$ vanishes for all loops $\alpha$, iff $\int_{\gamma} \vec{\bf G} \cdot d\vec{\bf s}$. In our example, we can take the loop $\gamma(t) := (\cos t, \sin t, 0)$, $0 \leq t \leq 2 \pi$, and computing gives
$$\int_{\alpha} \vec{\bf R} \cdot d\vec{\bf s} = 0 \qquad \textrm{but} \qquad \int_{\alpha} \vec{\bf V} \cdot d\vec{\bf s} = 2\pi ,$$
and we conclude that $\vec{\bf R}$ is conservative but $\vec{\bf V}$ is not, as claimed.
