Let $f(r)$ be a scalar function from $\Bbb R \mapsto \Bbb R$. Does the Vector field $\vec{F}(\vec{r})=\vec{r}f(r)$ have a potential? If yes, determine the potential function.
$$\vec{F}(\vec{r})=\begin{pmatrix}xy\\yz\\zx\end{pmatrix}$$
I have been trying to figure this out by using the fact that:
$$\vec{F} \space \space\text{is conservative} \iff \vec{F}=\nabla f(r)$$
but I am not really getting anywhere. Any ideas?
You can check that $\vec F(\vec r)=\vec rf(r)$ is actually the equation of a central force and all central forces are conservative. If you do $\nabla \times \vec F$, you will find that the curl is $0$.
So, we can say that there exists a $\phi$ such that $\vec F=\nabla \phi$. Now equating we have $$\nabla \phi=\vec rf(r)$$
$$\hat i\frac{\partial \phi}{\partial x}+\hat j\frac{\partial \phi}{\partial y}+\hat k\frac{\partial \phi}{\partial z}=\vec rf(r)$$
$$\hat i\frac{\partial \phi}{\partial r}\frac{\partial r}{\partial x}+\hat j\frac{\partial \phi}{\partial r}\frac{\partial r}{\partial y}+\hat k\frac{\partial \phi}{\partial r}\frac{\partial r}{\partial z}=\vec rf(r)$$
$$\hat i\frac{\partial \phi}{\partial r}\frac{x}{r}+\hat j\frac{\partial \phi}{\partial r}\frac{y}{r}+\hat k\frac{\partial \phi}{\partial r}\frac{z}{r}=\vec rf(r)$$ since $r^2=x^2+y^2+z^2$ and $r\frac{\partial r}{\partial x}=x$ and $r\frac{\partial r}{\partial y}=y$ and $r\frac{\partial r}{\partial z}=z$
$$\frac{\partial \phi}{\partial r}\frac{1}{r}\left(\hat i x+\hat jy+\hat kz\right)=\vec rf(r)$$
$$\frac{\partial \phi}{\partial r}\frac{\vec r}{r}=\vec rf(r)$$
Hence $\frac{\partial \phi}{\partial r}=rf(r)$ $$\phi= \int rf(r)dr = r\int f(r)dr- \int (\frac{d}{dr}(r))\int f(r)dr) dr$$ Therefore, $\phi = r\int f(r)dr- \int [\int f(r)dr] dr$ which depends on $f(r)$.