Find $\nabla \cdot (f\textbf r)$ and $\nabla \times (f\textbf r)$ of the function $f(x,y,z) = (x^2+y^2)\log(1-z)$ I have been given the function $f(x,y,z) = (x^2+y^2)\log(1-z)$ and I need to find the divergence $\nabla \cdot (f\textbf r)$ and curl $\nabla \times (f\textbf r)$ where $\textbf r$ is the position vector.
I understand that $$\nabla \cdot F = \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}$$
and that $$\nabla \times F = \begin{bmatrix}
                                i & j & k \\
                                \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\
                                F_x & F_y & F_z \\
                            \end{bmatrix}
                            $$
I just don't know how to proceed with this question.
 A: You multiply the function by the position vector and then just calculate the derivatives...
[\begin{array}{l}
\vec r = \left( {\begin{array}{*{20}{c}}
x\\
y\\
z
\end{array}} \right)\\
f(x,y,z) = \left( {{x^2} + {y^2}} \right)\log (1 - z) = {x^2}\log (1 - z) + {y^2}\log (1 - z)\\
f(x,y,z)\vec r = \vec F = \left( {\begin{array}{*{20}{c}}
{{x^3}\log (1 - z) + x{y^2}\log (1 - z)}\\
{{x^2}y\log (1 - z) + {y^3}\log (1 - z)}\\
{{x^2}z\log (1 - z) + {y^2}z\log (1 - z)}
\end{array}} \right)\\
\nabla  \bullet \vec F = {\partial _x}{F_x} + {\partial _y}{F_y} + {\partial _z}{F_z} = ...\\
\nabla  \times \vec F = \left| {\begin{array}{*{20}{c}}
{\hat i}&{\hat j}&{\hat k}\\
{{\partial _x}}&{{\partial _y}}&{{\partial _z}}\\
{{F_x}}&{{F_y}}&{{F_z}}
\end{array}} \right| = \left( {\begin{array}{*{20}{c}}
{{\partial _y}{F_z} - {\partial _z}{F_y}}\\
{{\partial _z}{F_x} - {\partial _x}{F_z}}\\
{{\partial _x}{F_y} - {\partial _y}{F_x}}
\end{array}} \right) = ...
\end{array}]
A: I think the easiest way to follow through with the calculations in a case like this is to use the well-known vector calculus identities,
$\nabla \cdot (fX) = \nabla f \cdot X + f \nabla \cdot X, \tag{1}$
$\nabla \times (fX) = \nabla f \times X + f \nabla \times X, \tag{2}$
where $f$ is a scalar function and $X$ is a vector field, both assumed to be sufficiently differentiable.
We also have, for $\mathbf r = (x, y, z)^T$, 
$\nabla \cdot \mathbf r = 3, \tag{3}$
and
$\nabla \times \mathbf r = 0; \tag{4}$
the first of these, (3), is very east to calculate; the second, (4), follows from the fact that the $x$ component of $\mathbf r$, $x$, exhibits no dependence on $y$ or $z$ and so forth; see my answer to this question.  Now with $f(x, y, z)$ as in the text of the present problem,
$f(x, y, z) = (x^2+y^2) \log(1-z), \tag{5}$
we have
$\nabla f = \begin{pmatrix} 2x \log(1-z) \\ 2y \log(1-z) \\ -\dfrac{x^2 + y^2}{1 - z} \end{pmatrix}; \tag{6}$
it follows from (1) and (3) that
$\nabla  \cdot (f \mathbf r) = \nabla f \cdot \mathbf r + 3 f$
$= 2x^2\log(1-z) + 2y^2 \log(1-z) - \dfrac{(x^2 + y^2)z}{1 - z} + 3(x^2 + y^2) \log(1 - z)$
$=  (5\log(1 - z) - \dfrac{z}{1 - z})(x^2 + y^2); \tag{7}$
as for $\nabla \times f \mathbf r$, from (2) and (4),
$\nabla \times f \mathbf r = \nabla f \times \mathbf r; \tag{9}$
in (9), we have reduced the problem to an ordinary vector cross product; I trust my readers can finish this calculation if they need to know the result.
