Definition: A vector $\bar{F}$ whose divergence $\bar{F}$ is zero is called solenoidal. For such a vector there is no loss or gain of fluid.

Given: $$\bar{F} = \frac{\bar{a} \times \bar{r}}{r^n} = \begin{vmatrix} i& j & k \\ a_{1} & a_{2} & a_{3} \\ \frac{x}{r^n} & \frac{y}{r^n} & \frac{z}{r^n} \end{vmatrix}$$

$$\bar{F} = i\bigg[\frac{a_{2}z}{r^n} - \frac{a_{3}y}{r^n}\bigg] - j\bigg[\frac{a_{1}z}{r^n} - \frac{a_{3}x}{r^n}\bigg] + k\bigg[\frac{a_{1}y}{r^n} - \frac{a_{3}x}{r^n}\bigg]$$

Now $$\bar{V} \cdot \bar{F} = \frac{\partial}{\partial x}\bigg[\frac{a_{2}z}{r^n} - \frac{a_{3}y}{r^n}\bigg] - \frac{\partial}{\partial y}\bigg[\frac{a_{1}z}{r^n} - \frac{a_{3}x}{r^n}\bigg] + \frac{\partial}{\partial z}\bigg[\frac{a_{1}y}{r^n} - \frac{a_{3}x}{r^n}\bigg]$$

$\therefore \frac{\partial}{\partial x}\bigg[\frac{a_{2}z}{r^n} - \frac{a_{3}y}{r^n}\bigg] = (a_{2}z - a_{3}y)\frac{\partial}{\partial x}r^{-n} \\ = (a_{2}z - a_{3}y)(-n)r^{n - 1} \frac{x}{r} [ \because \frac{\partial r}{\partial x} = \frac{x}{r}] \\ = (a_{2}z - a_{3}y) \frac{(-n)}{r^{n + 2}}x $

Similarly, we get

$$\frac{\partial}{\partial y}\bigg[\frac{a_{1}z}{r^n} - \frac{a_{3}x}{r^n}\bigg] = (a_{3}x - a_{1}z) \frac{(-n)}{r^{n + 2}}y$$

$$\frac{\partial}{\partial z}\bigg[\frac{a_{1}y}{r^n} - \frac{a_{3}x}{r^n}\bigg] = (a_{1}y - a_{2}x)\frac{(-n)}{r^{n + 2}}z$$

$$\therefore \bar{V} \cdot \bar{F} = \bigg[\frac{n}{r^{n + 2}}(a_{3}yx - a_{2}zx + a_{1}yz - a_{3}yx + a_{2}xz - a_{1}yz)\bigg] = 0$$

Hence $\bar{F}$ is solenoidal