I've been asked to do the following assignment.
Transform the following operations to cylindrical coordinates; let $A = \frac{7}{2} xi + 2y^2 j + 7zk$ and $B = 2xi + \frac{2}{3} j-2zk$ a) Divergence of A, b) double Rotational of A, c) rotational of A B.
Here is my approach:
For cylindrical coordinates
$$ x = \rho \ cos \ \phi \quad y = \rho \ sin \ \phi \ z = z$$
a) If I choose not to use general coordinates and just evaluate $\nabla A$ and then replace x,y and z from cylindrical coordinates gives
$$\nabla \cdot A = \frac {\partial A_1}{\partial \rho} + \frac {\partial A_2}{\partial \phi} + \frac {\partial A_3}{\partial z}$$
$$\nabla \cdot A = \frac {7x}{2} + 2 y^2 + 7z $$
Since x, y and z has a corresponding value for cylindrical coordinates replacing it gives:
$$\nabla \cdot A = \frac {\partial}{\partial \rho} \frac {7\rho \ cos \ \phi}{2} + \frac {\partial \ 2 \rho^2 sin^2 \ \phi}{\partial \phi} + \frac {\partial 7z}{\partial z}$$
$$\nabla \cdot A = \frac {7 cos \phi}{2} + 2 \rho^2 sin \ 2\phi + 7$$
Which I believe is the answer, however, If I do it by considering the general coordinates where:
$$ h_\rho = 1 \qquad h_\phi = \rho \qquad h_z = 1 $$
$$\nabla \cdot A = \frac {1}{h_\rho h_\phi h_z} [ \frac {\partial h_\phi h_z A_1 }{\partial \rho} + \frac {\partial h_z h_\rho A_2 }{\partial \phi} + \frac {\partial h_\rho h_\phi A_3 }{\partial z}]$$
Replacing h values and x,y and z in cylindrical coordinates
$$\nabla \cdot A = \frac {1}{\rho} [ \frac {\partial}{\partial \rho} \rho (\frac{7 \rho cos \ \phi}{2}) + \frac {\partial}{\partial \phi} (2 \rho^2 sin^2 \ \phi) + \frac {\partial}{\partial z} \rho(7z)]$$
$$\nabla \cdot A = \frac {1}{\rho} [ \frac {7 cos \phi}{2} (2 \rho) + 2\rho^2 (sin 2\phi) + 7\rho]$$
$$\nabla \cdot A = 7 \ cos \phi + 2\rho \ sin 2\phi + 7$$
As you can see I obtained two different result when doing $\nabla A$ when using general coordinates and just evaluating without using general coordinates. Which one is the correct answer? If you have any comments on my answer please let me know.
b) On this I don't know what it means double rotational, is it $\nabla^2 \times A$ ? or $\nabla \times (\nabla \times A)$
c) On this last one since I am confused on how to convert to cylindrical coordinates properly. If I try to do something like, let X be a vector and change it to cylindrical coordinates I know how to do it, but If I am asked to do something like evaluate:
$$ A \times B $$ or $$ A \cdot B $$ or $$ \nabla \cdot A $$
I do not understand how to do it (in exercise a I obtained two different answers). And last, what does it mean rotational of AB, is it $\nabla \times (A\cdot B)$ or is it $\nabla \times (A\times B)$.
Refer to this wiki page
Please note that $ \ \hat x, \hat y, \hat z$ unit vectors from cartesian coordinates are converted to cylindrical as, $\hat x = \cos \varphi \ \hat \rho - \sin \varphi \ \hat \varphi, \hat y = \sin \varphi \ \hat \rho + \cos \varphi \ \hat \varphi, \hat z = \hat z$.
So taking example of the first one, $\vec A = \frac{7}{2} x \hat i + 2y^2 \hat j + 7z \hat k$ can be written in cylindrical coordinates as,
$\vec A = \displaystyle \small \frac{7 \rho \cos \varphi}{2} (\cos \varphi \ \hat \rho - \sin \varphi \ \hat \varphi) + 2 \rho^2 \sin^2\varphi \ (\sin \varphi \ \hat \rho + \cos \varphi \ \hat \varphi) + 7z \ \hat z$
$ = \displaystyle \small \frac{7 \rho \cos^2 \varphi + 4 \rho^2 \sin^3\varphi}{2} \hat \rho + \frac{4 \rho^2 \sin^2 \varphi \cos \varphi - 7 \rho \sin\varphi \cos\varphi}{2} \hat \varphi + 7z \ \hat z$
Now in cylindrical coordinates, $\nabla \cdot \vec A = \displaystyle \small \frac{1}{\rho}\frac{\partial (\rho A_{\rho})}{\partial \rho} + \frac{1}{\rho}\frac{\partial A_{\varphi}}{\partial \varphi} + \frac{\partial A_{z}}{\partial z}$
The first term evaluates to $(7 \cos^2\varphi + 6 \rho \sin^2\varphi)$, the second term evaluates to $(4 \rho \sin \varphi \cos^2\varphi - 2 \rho\sin^3\varphi - \frac{7}{2}\cos2\varphi)$ and the last term is simply $7$.
Adding them up and simplifying, you get
$\nabla \cdot \vec A = \displaystyle \small \frac{21}{2} + 4 \rho \sin \varphi$
Can you take it from here?