2

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.

user145608
  • 99

2 Answers2

2

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}]

0

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.

Community
  • 1
Robert Lewis
  • 71,180