# Transforming from cartesian to cylindrical

Here is the question:

Transform $\textbf{A} = \hat{\mathbf{x}} 2 - \hat{\mathbf{y}}5 + \hat{\mathbf{z}}3$ into cylindrical coordinates at point ($x=-2, y=3, z=1$).

What I have tried is this: consider the point the exercise has given and find $r$ and $\theta$ with it. This leads to

$$r = \sqrt{(-2)^2 + 3^2} = \sqrt{13}, \quad \cos \theta = \frac{-2}{\sqrt{13}}, \sin \theta = \frac{3}{\sqrt{13}}.$$

The book gives the relation

$$\begin{bmatrix} A_r \\ A_{\theta} \\ A_z \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta & 0 \\ - \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} A_x \\ A_y \\ A_z \end{bmatrix}$$

as the relation of the coordinates. Using this, I plugged the values of $\cos \theta, \sin \theta$ I found and took $A_x = 2, A_y = -5, A_z = 3$. Therefore,

$$\begin{bmatrix} A_r \\ A_{\theta} \\ A_z \end{bmatrix} = \begin{bmatrix} 2 \cos \theta - 5 \sin \theta \\ - 2 \sin \theta -5 \cos \theta \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{2 \cdot (-2) - 5 \cdot (3)}{\sqrt{13}} \\ \frac{-2 \cdot (-2) -5 \cdot (3)}{\sqrt{13}} \\ 3 \end{bmatrix} = \begin{bmatrix} \frac{-19}{\sqrt{13}} \\ \frac{- 11}{\sqrt{13}} \\ 3 \end{bmatrix}.$$

This gives approximately $(A_r, A_{\theta}, A_z) = (- 5.27, 3.05, 3)$, but the answer is

$$\textbf{A} = \hat{\mathbf{r}} 5.27 - \hat{\mathbf{\theta}}1.11 + \hat{\mathbf{z}}3.$$

Seemingly I get a wrong sign for the first coordinate and the second one is completely off. What am I doing wrong (if so)?

This comes from self-studying electromagnetism using this book, first chapter (vector algebra).

When you substituted in for the trig functions in the $A_{\theta}$ line, you put the values in wrong. You should have $-2\sin \theta - 5 \cos \theta=\frac 1{\sqrt{13}}(2(-2)-5(-2))=\frac 4{\sqrt{13}}\approx 1.11$
Now we have an overall sign error relative to the book. As I draw the picture, the book is wrong. The given point is in the second quadrant, while $\textbf{A}$ points down and toward the origin. $A_r$ is therefore negative. $\textbf{A}$ also points to the left of the radial vector from the point to the origin, so $A_{\theta}$ is positive.
• Embarassing, as I thought I had it right when I recalculated the second or third time. Thanks! The sign error could be due to a orientation reversing coordinate change in the $xy$ plane, right? The $z$ axis would be essentially unaltered. – Mark Fantini Dec 19 '13 at 22:09
• The way I learned it, $A_r$ was the component of $\textbf A$ in the direction of the local radius, with positive oriented out. $A_{\theta}$ was the component of $\textbf A$ perpendicular to the local radius, positive when pointing counterclockwise. Those are the conventions I used above. – Ross Millikan Dec 19 '13 at 22:13