A vector field is spherically symmetric about the origin if, on every sphere centered at the
origin, it has constant magnitude and points either away from or toward the origin. A vector
field that is spherically symmetric about the origin can be written in terms of the spherical
coordinate $F = f(\rho)e_\rho$ where $f$ is a function of the distance $ρ$ from the origin,
$f(0) = 0$, and $e_ρ$ is a unit vector pointing away from the origin.Show that $\operatorname{div} F = \frac 1 {\rho^2} \frac{d(\rho^2f(\rho))}{d \rho}
, ρ \neq 0.$Source: Hughes-Hallet, freely available here (project #2).
My work is below. I request verification or feedback.
Note: This problem is trivial to solve using the Divergence Theorem in Spherical Coordinates. But that has not yet been taught in the book; the point of this problem is to derive Divergence Theorem in Spherical Coordinates, or at least as much as you need to solve the problem.
Consider the point $v_0$ on the $x$ axis such that $$\begin{align*}v_0 &= \langle \rho_0, 0, 0 \rangle\\
f_0 &= f(\rho_0)\\
F(v_0) &= \frac {f_0}{\rho_0}v_0.\end{align*}$$ Taking a step $\partial x$ in the $x$ direction doesn't change the direction of $v$, but only increases $\rho$, giving $$\begin{align*}
v_1 &= \langle \rho_0 + \partial x, 0, 0 \rangle \\
f_1 &= f(\rho_0 + \partial x) \\
F(v_1) &= \frac {f_1}{\rho_0}v_0 \\
\frac{\partial F_x}{\partial x} &= \frac{\partial f}{\partial \rho}.
\end{align*}$$
Now, instead of stepping in the $x$ direction, consider starting from $v_0$ and taking a step $\partial y$ in the $y$ direction: $$v_2 = \langle \rho_0, \partial y, 0 \rangle.$$ Since this step is orthogonal to $v_0$, $$\rho_2 = \rho_0 + \mathcal{O}([\partial y]^2)$$ and since $f$ depends only on $\rho$, we have to a first order approximation $$f_2 \approx f_0 \\
F(v_2) \approx \frac {f_0}{\rho_0}v_2$$ and, in the limit, $$\frac{\partial F_y}{\partial y} = \frac f \rho.$$
By symmetry, $\frac{\partial F_z}{\partial z} = \frac f \rho$ as well, giving $$\begin{align*}
\operatorname{div} F &= \frac{\partial f}{\partial \rho} + 2\frac f \rho \\
&= \frac 1 {\rho^2} \frac{d(\rho^2f(\rho))}{d \rho}
\end{align*}$$ as desired.
The above applies directly to points on the $x$ axis (other than the origin), and, by symmetry to any point in space (besides the origin).
Best Answer
OK, by your request, here's how I would present your heuristic argument.
It is important to observe that the divergence operator is rotationally invariant (see this brief justification). Since $F$ is invariant under rotation about the origin (being both rotationally symmetric in magnitude and radial), it follows that we can make the analysis at a single point and deduce that it holds at all points on the sphere of that radius.
So, as you did in the OP, let's consider $F=f(\rho)e_\rho$ at the point $a=(x,0,0)$ with $x>0$. Let's write $F=(M,N,P)$ for simplicity. Then $M(a)=f(x)$ and $N(a)=P(a)=0$.
Then varying $x$ is varying $\rho$ and keeping $\phi$ and $\theta$ fixed. It follows that $$\frac{\partial M}{\partial x}(a) = \frac{\partial M}{\partial\rho}\Big|_{\rho=x,\phi=\theta=0} = f'(x).$$
Now, what about $\dfrac{\partial N}{\partial y}(a)$ and $\dfrac{\partial P}{\partial z}(a)$? As we move from the point $a$ varying $y$ (or $z$), instantaneously the point moves tangent to the sphere of radius $x$ centered at the origin, so instantaneously $\rho$ does not vary as we move in the $y$ direction. However, we must remember that we have $e_\rho =(x,y,z)/\rho$, so $N= f(\rho)\dfrac y\rho$. At the point $a$, we have $\dfrac{\partial\rho}{\partial y} = \dfrac y\rho = 0$. It follows from the product rule and chain rule that
$$\frac{\partial N}{\partial y}(a) = f’(x)\frac{\partial \rho}{\partial y}\Big|_a + f(x)\cdot\frac1\rho\Big|_a + f(x)\cdot 0\cdot\frac{\partial \rho}{\partial y}\Big|_a = \frac{f(\rho)}{\rho}\Big|_{\rho=x}.$$ Similarly for $(\partial P/\partial z)(a)$.
We conclude that $\text{div}\,F(a) = \dfrac{\partial M}{\partial x}(a) + \dfrac{\partial N}{\partial y}(a) + \dfrac{\partial P}{\partial z}(a) = f'(\rho) + 2\dfrac{f(\rho)}\rho$, as desired.