Answer
$\boxed{\text{div }\mathbf{F} = \frac{2}{\sqrt{x^2+y^2+z^2}}, \quad
\text{curl }\mathbf{F} = \mathbf{0}}$
Work Step by Step
Let
$\mathbf{F}(x,y,z) = \frac{x \mathbf{i} + y \mathbf{j} + z \mathbf{k}}{\sqrt{x^2+y^2+z^2}}.$
Divergence of $\mathbf{F}$
The divergence is defined as
$\text{div }\mathbf{F} = \nabla \cdot \mathbf{F} = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}.$
Here,
$F_x = \frac{x}{\sqrt{x^2+y^2+z^2}},$
$F_y = \frac{y}{\sqrt{x^2+y^2+z^2}}, $
$F_z = \frac{z}{\sqrt{x^2+y^2+z^2}}.$
Compute the partial derivatives:
$\frac{\partial F_x}{\partial x} = \frac{\partial}{\partial x} \left( \frac{x}{\sqrt{x^2+y^2+z^2}} \right)
= \frac{\sqrt{x^2+y^2+z^2} - \frac{x \cdot x}{\sqrt{x^2+y^2+z^2}}}{x^2+y^2+z^2}
= \frac{y^2+z^2}{(x^2+y^2+z^2)^{3/2}}.$
Similarly,
$\frac{\partial F_y}{\partial y} = \frac{x^2+z^2}{(x^2+y^2+z^2)^{3/2}}, \quad
\frac{\partial F_z}{\partial z} = \frac{x^2+y^2}{(x^2+y^2+z^2)^{3/2}}.$
Thus,
$\text{div }\mathbf{F} = \frac{y^2+z^2 + x^2+z^2 + x^2+y^2}{(x^2+y^2+z^2)^{3/2}}
= \frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{3/2}}
= \frac{2}{\sqrt{x^2+y^2+z^2}}.$
Curl of $ \mathbf{F}$
The curl is defined as
$\text{curl }\mathbf{F} = \nabla \times \mathbf{F} =$
$\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
\frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
F_x & F_y & F_z
\end{vmatrix}.$
Since $F_x, F_y, F_z$ are proportional to $ x, y, z$ respectively, and the field is radial,
$\text{curl }\mathbf{F} = \mathbf{0}.$
$\boxed{\text{div }\mathbf{F} = \frac{2}{\sqrt{x^2+y^2+z^2}}, \quad
\text{curl }\mathbf{F} = \mathbf{0}}$
