<s>[INST] Finding all values of $p$ for which $\operatorname{div}\vec{F} = 0$ Here is the full question:
Let $\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$ and let $r = \|\vec{r}\|$. Let $\vec{F} = r^p\vec{r}$. Find all values of $p$ for which div $\vec{F} = 0$
I'm a bit confused on this as it seems that at any point in the vector field, $r^p$ cannot equal zero for any value of $p$ unless $r = 0$, then the value of $p$ would be irrelevant.
So, if I were to start finding div $\vec{F}$, the first thing I would need to do it find:
$$\operatorname{div}\vec{F} = \frac{\partial}{\partial x} \left(x\sqrt{x^2+y^2+z^2}^p\right) + \frac{\partial F_2}{\partial x} + \frac{\partial F_3}{\partial x}$$
And looking just at the first partial:
$$\frac{\partial}{\partial x}\left(x\sqrt{x^2+y^2+z^2}^p\right) = 2x^2\left(p\sqrt{x^2+y^2+z^2}^{p-1}\right) + \sqrt{x^2+y^2+z^2}^p$$
In order to get div $\vec{F} = 0$, $F_1$, $F_2$, and $F_3$ must go to zero when taking the partials or be zero, however, if I did the above correctly, they will not go to zero, regardless of the value of $p$.
Am I completely missing the mark here?
Thank you very much for your time and assistance.
Best,
Eric
 [/INST] Observe that in taking one of the partial derivatives we find that 
\begin{eqnarray*}
\frac{\partial}{\partial x} x \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}} & = & \frac{p}{2} x \cdot (2x) \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}-1} + \left (x^2 + y^2 + z^2 
\right )^{\frac{p}{2}} \\
& = & px^2 \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}-1} + ||\textbf{r}||^p.
\end{eqnarray*}
When we sum all three components we find that 
\begin{eqnarray*}
\nabla \cdot \textbf{F} & = & p(x^2 + y^2 + z^2) \left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}-1} + 3||\textbf{r}||^p \\
& = & p\left (x^2 + y^2 + z^2 \right )^{\frac{p}{2}} + 3||\textbf{r}||^p \\
& = & p||\textbf{r}||^p + 3||\textbf{r}||^p
\end{eqnarray*}
and this quantity equals zero only when $p = -3$.
</s>