Why is $\nabla \times (f(r) \vec r) = 0$?

Question

With $r=\lvert \vec r\rvert$

I know how to work with $\nabla$ , but I don't know how to deal with $f(r) \vec r$ ... Can you help me?

Answer 1

This is just a computation. Try directly, as example I calculate the firt component:

$$(\nabla\times f(r)\vec r)_x= \displaystyle\frac{\partial}{\partial y}f(r)z-\displaystyle\frac{\partial}{\partial z}f(r)y $$

since

$$\displaystyle\frac{\partial}{\partial y}f(r)=f'(r)\displaystyle\frac{y}{r} $$

and similar

$$\displaystyle\frac{\partial}{\partial z}f(r)=f'(r)\displaystyle\frac{z}{r} $$

then

$$(\nabla f(r)\vec r)_x=f'(r)\displaystyle\frac{zy}{r}-f'(r)\displaystyle\frac{yz}{r}=0$$

In the same way you can compute the others components.