How to prove in two ways that a 3d vector field is conservative?

Question

Let $\mathbf{r}=\langle x,y,z\rangle$ be a vector. It's length is $r=\sqrt{x^2+y^2+z^2}$. Let $$\mathbf{F}=\frac{\mathbf r}{r}$$ be a vector field.

Prove in two ways that $\mathbf{F}$ is conservative over the set $x^2+y^2+z^2\ge1$.

The first and the easiest way is to compute $\operatorname{curl}\mathbf F$ which is zero.

The second way I thought of is that if a field is conservative then $\oint_C\mathbf F\cdot d\mathbf r=0$ over curve $C$.

I'm not sure what the curve is. I suppose that it's of form: $$ c(t)=\langle x,y, z\rangle $$ Because our surface is a sphere then $||c(t)||=k$ where $k$ is a constant which is essentially the radius of the sphere. If $||c(t)||=k$ then $c(t)\cdot c'(t)=0$ therefore: $$ \oint_C\mathbf F\cdot d\mathbf r=0 $$ Am I missing something? Is my proof fine?

Answer 1

Do you see that the vector field can be rewritten as $$\mathbf{F}=\frac{\mathbf{r}}{r}=-\nabla\Big(\frac{1}{r}\Big)$$ Define $\phi(r)=\frac{1}{r}$ and observe $$\oint_C\mathbf{F}\cdot d\mathbf{r}=-\oint_C\nabla\phi\cdot d\mathbf{r}=0$$