Mean square radius vs mean radius in nuclear physics There are several ways to define the radius of a nucleus, one of which is the mean square radius:
$$ \langle r^2 \rangle  = \frac{\int r^2 \rho (r) d\tau}{\int \rho (r) d\tau}$$
What is the convenience of this opposed to:
$$ \langle r \rangle = \frac{\int r\rho(r)d\tau}{\int \rho(r)d\tau}$$
 A: In principle you can define the nuclear radius in many ways, but we have a preference for definitions that can be related to experiment in a straightforward way. First of all, notice that text books typically discuss electric charge radii, that means $\rho(r)$ is the density of electric charge, not the baryon density. This is a short coming, because $\rho(r)$ is not sensitive to neutrons, but it has the advantage that the electric charge distribution is easily probed by electron scattering. In the the low energy limit electron scattering simply measures the charge form factor
$$
 F(q^2) = \int dr^3 e^{i\vec{q}\cdot\vec{r}}\rho(\vec{r})
$$ 
For small $q$ I can expand this out. Because of rotational invariance there is no linear term, and the leading correction to the total charge $F(0)=Q$ is the slope of the form factor
$$
F'(0) =  -\frac{1}{6}\int dr^3 r^2\rho(\vec{r})
$$
which is related to the mean square radius. 
A: We consider the nuclear charge distribution to be uniform i.e., $$\rho(r)=Constant$$
Also, Elmentary volume in spherical-polar co-ordiantes
$$d\tau= r^2 \sin\theta drd\theta d\phi$$
Now,    $$\left< r^2 \right> =\frac{\iiint r^2 \rho(r)d\tau}{\iiint \rho(r)d\tau}$$
$$\left< r^2 \right> =\frac{\iiint r^2\rho(r) \times r^2 \sin\theta drd\theta d\phi}{\iiint r^2 \rho(r) \sin\theta drd\theta d\phi}$$
$$r:0\to R$$
$$\theta:0\to \pi$$
$$\phi : 0 \to 2\pi$$
$$\left< r^2 \right> =\frac{\int_0^R r^4 dr}{\int_0^R r^2 dr}$$
$$\left< r^2 \right> =\frac{R^5/5}{R^3/3}=\frac{3R^2}{5}$$
Similarly, $$\left< r \right> =\frac{\iiint r\rho(r) \times r^2 \sin\theta drd\theta d\phi}{\iiint r^2\rho(r) \sin\theta drd\theta d\phi}$$
$$\left< r \right> =\frac{\int_0^R r^3dr}{\int_0^R r^2 dr}=\frac{R^4/4}{R^3/3}=\frac{3R}{4}$$
