# Transform a vector from Cartesian to Cylindrical coordinates

1. Sep 4, 2008

### VinnyCee

1. The problem statement, all variables and given/known data

Transform the vector below from Cartesian to Cylindrical coordinates:

$$Q\,=\,\frac{\sqrt{x^2\,+\,y^2}}{\sqrt{x^2\,+\,y^2\,+\,z^2}}\,\hat{x}\,-\,\frac{y\,z}{x^2\,+\,y^2\,+\,z^2}\,\hat{z}$$

2. Relevant equations

Use these equations:

$$A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A}\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi$$

$$A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi$$

$$A_z\,=\,A_z$$

$$x\,=\,\rho\,cos\,\phi$$

$$y\,=\,\rho\,sin\,\phi$$

3. The attempt at a solution

$$Q_{\rho}\,=\,\frac{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2}}{\sqrt{\left(\rho\,cos\,\phi\right)^2\,+\,\left(\rho\,\sin\,\phi\right)^2\,+\,z^2}}\,cos\,\phi$$

$$Q_{\rho}\,=\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,cos\,\phi$$

$$Q_{\phi}\,=\,-\,\frac{\rho}{\sqrt{\rho^2\,+\,z^2}}\,sin\,\phi$$

$$Q_z\,=\,-\,\frac{z\,\rho\,sin\,\phi}{\rho^2\,+\,z^2}$$

Does the above look correct?

Last edited: Sep 4, 2008
2. Sep 5, 2008

### tiny-tim

Hi VinnyCee!
No … these are the equations for an ordinary rotation of the x and y axes through an angle phi.
hmm … you want polar coordinates of a vector along the x-axis … so what will its phi coordinate be?

3. Sep 5, 2008

### VinnyCee

Really? The equations for transforming from Cartesian to Cylindrical were given in the notes for the class in a matrix form and then solved out to a formula, the one below.

Here is a more detailed version for the $\rho$ coordinate transformation only:

$$A_{\rho}\,=\,\hat{\rho}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\rho}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\rho}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\rho}\,\cdot\,\hat{z}\right)\,A_z\,=\,A_x\,cos\,\phi\,+\,A_y\,sin\,\phi$$

Is that incorrect? If so, what is the correct way to transform from Cartesian to Cylindrical?

I think the $\phi$ coordinate would be zero in this case since there is no y-component, right?

4. Sep 5, 2008

### Redbelly98

Staff Emeritus

Here you've included the Ax cos(phi) term, but forgot the Ay sin(phi) term

5. Sep 5, 2008

### VinnyCee

$A_y$ is zero, so it need not be included, right?

6. Sep 5, 2008

### Redbelly98

Staff Emeritus
Oops, I was confusing Ay and Az. My bad.

Your results from post #1 look okay to me. My understanding is the dot product of the vector Q with the coordinate unit-vector is the component of Q in that coordinate direction, which is exactly what you did.

On the other hand I'm hesitant to disagree with tiny-tim, since he knows this stuff pretty well also.

7. Sep 5, 2008

### tiny-tim

That's right! for x > 0 and y = 0, phi must be zero.
No!

A three-dimensional vector needs three coordinates …

if one coordinate is 0, you have to say so!
No, that is correct … the other one is wrong.

8. Sep 5, 2008

### Redbelly98

Staff Emeritus
Isn't that the same expression as in post #1 for Aρ ?

Last edited: Sep 5, 2008
9. Sep 5, 2008

### tiny-tim

ah … I meant the phi-coordinate $$A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A}\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi$$ is wrong.

10. Sep 5, 2008

### Redbelly98

Staff Emeritus
Um, if we use the convention that φ's unit vector points counterclockwise, and φ is taken counterclockwise from the positive x-axis, then I agree with VinnyCee's formula.

11. Sep 5, 2008

### VinnyCee

So, does that mean that the original post contains the correct answer? Or is the $Q_{\phi}$ variable still going to be zero even though the formulas given produce the expression I wrote in the first post?

The expression written from a matrix in class is:

$$A_{\phi}\,=\,\hat{\phi}\,\cdot\,\overrightarrow{A} \,=\,\left(\hat{\phi}\,\cdot\,\hat{x}\right)\,A_x\,+\,\left(\hat{\phi}\,\cdot\,\hat{y}\right)\,A_y\,+\,\left(\hat{\phi}\,\cdot\,\hat{z}\right)\,A_z\,=\,-A_x\,sin\,\phi\,+\,A_y\,cos\,\phi$$

Since...

$$\hat{x}\,\cdot\,\hat{\phi}\,=\,-\,sin\,\phi$$

$$\hat{y}\,\cdot\,\hat{\phi}\,=\,cos\,\phi$$

$$\hat{z}\,\cdot\,\hat{\phi}\,=\,0$$

Right?

Last edited: Sep 5, 2008
12. Sep 6, 2008

### tiny-tim

sorry!

Hi VinnyCee and Redbelly98!

Yes, I completely mis-read the original question.