3 Cylindrical polar coordinates

In cylindrical polar coordinates $\left(\rho ,\varphi ,z\right)$ , the three unit vectors are $\underset{̲}{\stackrel{̂}{\rho }}$ , $\underset{̲}{\stackrel{̂}{\varphi }}$ and $\underset{̲}{ẑ}$ (see Figure 20(b) on page 38) with scale factors

$h_{\rho }=1$ , $h_{\varphi }=\rho$ , $h_z=1$ .

The quantities $\rho$ and $\varphi$ are related to $x$ and $y$ by $x=\rho cos\varphi$ and $y=\rho sin\varphi$ . The unit vectors are $\underset{̲}{\stackrel{̂}{\rho }}=cos\varphi \underset{̲}{i}+sin\varphi \underset{̲}{j}$ and $\underset{̲}{\stackrel{̂}{\varphi }}=-sin\varphi \underset{̲}{i}+cos\varphi \underset{̲}{j}$ . In cylindrical polar coordinates,

grad $f=\underset{̲}{\nabla }f=\frac{\partial f}{\partial \rho }\underset{̲}{\stackrel{̂}{\rho }}+\frac{1}{\rho }\frac{\partial f}{\partial \varphi }\underset{̲}{\stackrel{̂}{\varphi }}+\frac{\partial f}{\partial z}\underset{̲}{ẑ}$

The scale factor $\rho$ is necessary in the $\varphi$ -component because the derivatives with respect to $\varphi$ are distorted by the distance from the axis $\rho =0$ . If $\underset{̲}{F}=F_{\rho }\underset{̲}{\stackrel{̂}{\rho }}+F_{\varphi }\underset{̲}{\stackrel{̂}{\varphi }}+F_z\underset{̲}{ẑ}$ then

div $\underset{̲}{F}=\underset{̲}{\nabla }\cdot \underset{̲}{F}=\frac{1}{\rho }\left[\frac{\partial }{\partial \rho }\left(\rho F_{\rho }\right)+\frac{\partial }{\partial \varphi }\left(F_{\varphi }\right)+\frac{\partial }{\partial z}\left(\rho F_z\right)\right]$

curl $\underset{̲}{F}=\underset{̲}{\nabla }\times \underset{̲}{F}=\frac{1}{\rho }\left|\begin{array}{ccc}\hfill \underset{̲}{\stackrel{̂}{\rho }}\hfill & \hfill \rho \underset{̲}{\stackrel{̂}{\varphi }}\hfill & \hfill \underset{̲}{ẑ}\hfill \\ \hfill \hfill \\ \hfill \frac{\partial }{\partial \rho }\hfill & \hfill \frac{\partial }{\partial \varphi }\hfill & \hfill \frac{\partial }{\partial z}\hfill \\ \hfill \hfill \\ \hfill F_{\rho }\hfill & \hfill \rho F_{\varphi }\hfill & \hfill F_z\hfill \end{array}\right|$ .

Example 20

Working in cylindrical polar coordinates, find $\underset{̲}{\nabla }f$ for $f={\rho }^2+z^2$

Solution

If $f={\rho }^2+z^2$ then $\frac{\partial f}{\partial \rho }=2\rho$ , $\frac{\partial f}{\partial \varphi }=0$ and $\frac{\partial f}{\partial z}=2z$ so $\underset{̲}{\nabla }f=2\rho \underset{̲}{\stackrel{̂}{\rho }}+2z\underset{̲}{ẑ}$ .

Example 21

Working in cylindrical polar coordinates find

1. $\underset{̲}{\nabla }f$ for $f={\rho }^{3}sin\varphi$
2. Show that the result for 1. is consistent with that found working in Cartesian coordinates.

Solution

1. If $f={\rho }^{3}sin\varphi$ then $\frac{\partial f}{\partial \rho }=3{\rho }^{2}sin\varphi$ , $\frac{\partial f}{\partial \varphi }={\rho }^{3}cos\varphi$ and $\frac{\partial f}{\partial z}=0$ and hence,

   $\underset{̲}{\nabla }f=3{\rho }^{2}sin\varphi \underset{̲}{\stackrel{̂}{\rho }}+{\rho }^{2}cos\varphi \underset{̲}{\stackrel{̂}{\varphi }}$ .

2. $f={\rho }^{3}sin\varphi ={\rho }^2\rho sin\varphi =\left(x^2+y^2\right)y=x^{2}y+y^3$ so $\underset{̲}{\nabla }f=2xy\underset{̲}{i}+\left(x^2+3y^2\right)\underset{̲}{j}$ .
   
   Using cylindrical polar coordinates, from 1. we have $$\begin{array}{rcll}\underset{̲}{\nabla }f& =& 3{\rho }^{2}sin\varphi \underset{̲}{\stackrel{̂}{\rho }}+{\rho }^{3}cos\varphi \underset{̲}{\stackrel{̂}{\varphi }}& \\ & =& 3{\rho }^{2}sin\varphi \left(cos\varphi \underset{̲}{i}+sin\varphi \underset{̲}{j}\right)+{\rho }^{2}cos\varphi \left(-sin\varphi \underset{̲}{i}+cos\varphi \underset{̲}{j}\right)& \\ & =& \left[3{\rho }^{2}sin\varphi cos\varphi -{\rho }^{2}sin\varphi cos\varphi \right]\underset{̲}{i}+\left[3{\rho }^2{sin}^2\varphi +{\rho }^2{cos}^2\varphi \right]\underset{̲}{j}& \\ & =& \left[2{\rho }^{2}sin\varphi cos\varphi \right]\underset{̲}{i}+\left[3{\rho }^2{sin}^2\varphi +{\rho }^2{cos}^2\varphi \right]\underset{̲}{j}=2xy\underset{̲}{i}+\left(3y^2+x^2\right)\underset{̲}{j}& \end{array}$$

   So the results using Cartesian and cylindrical polar coordinates are consistent.

Example 22

Find $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ for $\underset{̲}{F}=F_{\rho }\underset{̲}{\stackrel{̂}{\rho }}+F_{\varphi }\underset{̲}{\stackrel{̂}{\varphi }}+F_z\underset{̲}{ẑ}={\rho }^3\underset{̲}{\stackrel{̂}{\rho }}+\rho z\underset{̲}{\stackrel{̂}{\varphi }}+\rho zsin\varphi \underset{̲}{ẑ}$ . Show that the results are consistent with those found using Cartesian coordinates.

Solution

Here, $F_{\rho }={\rho }^3$ , $F_{\varphi }=\rho z$ and $F_z=\rho zsin\varphi$ so

Converting to Cartesian coordinates,

So

So $\underset{̲}{\nabla }\cdot \underset{̲}{F}$ is the same in both coordinate systems.

Example 23

Find $\underset{̲}{\nabla }\times \underset{̲}{F}$ for $\underset{̲}{F}={\rho }^2\underset{̲}{\stackrel{̂}{\rho }}+zsin\varphi \underset{̲}{\stackrel{̂}{\varphi }}+2zcos\varphi \underset{̲}{ẑ}$ .

Solution