homework and exercises - Radially symmetric charge distribution (dipole moment)

a) There's a radially symmetric charge density $\rho(r)$ centered around the origin. Determine the dipolemoment of that charge density.

b) Let $\rho(r)$ be an arbitrary charge density now. Under what circumstances does the dipole moment of the displaced charge density $\rho '(\vec{r}) = \rho (\vec{r}-\vec{b})$ differ from the one not displaced at all.

Here were my ideas so far:

a) Just thinking about the situation it has to be zero, right? I mean, since there's no real dipole. But how do I show that mathematically?

I was thinking of just going like this (it may be wrong):

Let the charge density be $\rho (r)=kr$, then we can get the charge q by integrating:

$$q=4\pi \int_0^R kr\cdot r^2dr=\pi k R^4$$

I'm looking at the charge distribution as a spherical electron cloud with radius $R$.

Then, since $p=qd$ and $d$ is zero because there are no two different charges the dipole moment is zero. Is that sufficient as an answer?

b) I don't know how to approach this one. My guess is that its dipole moment is also zero because we're only looking at a displacement here.

Anyone got any idea? I would appreciate any advice on this.