I have a question regarding this topic:
$\vec A = \frac {3 \vec r}{r^2}$ is our vector field. We need to find the flux for the enclosed volume of a sphere with radius R, that has a parametrized surface $\partial V=(R \cos\phi \sin \theta,R \sin\phi \sin \theta,R \cos \theta)$.
$\vec A = 3(\frac{x}{r^2},\frac{y}{r^2},\frac{z}{r^2})$
Now the general parametrisation of $\vec r$ is $\vec r=(\rho \cos\phi \sin \theta,\rho \sin\phi \sin \theta,\rho \cos \theta))$.
Then : $\vec A=3(\frac{\rho \cos\phi \sin \theta}{\rho^2},\frac{\rho \sin\phi \sin \theta}{\rho^2},\frac{\rho \cos\theta}{\rho^2})$.
If in this expression:
$$\int_{\partial V}\vec A(\vec r) d \vec f=\int_{\partial V}\vec A(\rho,\phi,\theta)$$
where $d\vec f$ is an infinitesimal surface element, we plug in the corresponding values, the result is not the same with the result when we write $\vec A$ as:
$\vec A=3(\frac{ \cos\phi \sin \theta}{R},\frac{\sin\phi \sin \theta}{R},\frac{ \cos\theta}{R})$ which gives you a result of $12\pi R$.
But my question is, why do we write $\vec A=3(\frac{ \cos\phi \sin \theta}{R},\frac{\sin\phi \sin \theta}{R},\frac{ \cos\theta}{R})$ when clearly $\vec r$ can take arbitrary values so that $|\vec r|< R$ or $|\vec r|> R$ or $|\vec r|=R$
