Converting vector in cartesian to cylindrical coordinates

[dingo_d]

Homework Statement

This seems like a trivial question (because it is), and I'm just not sure if I'm doing it right.

I have vector in cartesian coordinate system:

$$\vec{a}=2y\vec{i}-z\vec{j}+3x\vec{k}$$

And I need to represent it in cylindrical and spherical coord. system

Homework Equations

$$a_\rho=a_x\cos\phi+a_y\sin\phi$$
$$a_\phi=-a_x\sin\phi+a_y\cos\phi$$
$$a_z=a_z$$

The Attempt at a Solution

What is cofusing me is this:
The formula for $$\phi$$ is$$\phi=\arctan\frac{y}{x}$$. Are those x and y in fact $$a_x$$ and $$a_y$$?

By some kind of reasoning it should be. But then $$\phi$$ is $$\phi=\arctan\frac{-z}{2y}$$ :\

Is this correct? And do I need to change the unit vectors too?

 

[planck42]

You can represent the $$\phi$$-component of a cylindrical/spherical vector in terms of $$\phi$$, like how you can represent the x-component of a Cartesian vector in terms of x. $${\phi}=tan^{-1}\frac{y}{x}$$ doesn't refer to the components of a vector [field]. Finally, unit vectors change according to the Jacobian matrix e.g. the transformation from the x unit vector to the $$\rho$$ unit vector is the x$$\rho$$-component of the Jacobian, or $$x_{\rho}$$

 

[dingo_d]

Ok so you say I should use Jacobian, which is

$$J(r,\phi, z)=\begin{bmatrix} {dx\over dr} & {dx\over d\phi} &{dx\over dz} \\ {dy\over dr} & {dy\over d\phi} & {dy\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} {d(r\cos\phi)\over dr} & {d(r\cos\phi)\over d\phi} & {d(r\cos\phi)\over dz} \\ {d(r\sin\phi)\over dr} & {d(r\sin\phi)\over d\phi} & {d(r\sin\phi)\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

But still where do I get the $$\phi$$? :\

 

[planck42]

Ok so you say I should use Jacobian, which is

$$J(r,\phi, z)=\begin{bmatrix} {dx\over dr} & {dx\over d\phi} &{dx\over dz} \\ {dy\over dr} & {dy\over d\phi} & {dy\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} {d(r\cos\phi)\over dr} & {d(r\cos\phi)\over d\phi} & {d(r\cos\phi)\over dz} \\ {d(r\sin\phi)\over dr} & {d(r\sin\phi)\over d\phi} & {d(r\sin\phi)\over dz} \\ {dz\over dr} & {dz\over d\phi} & {dz\over dz}\end{bmatrix} =\begin{bmatrix} \cos\phi & -r\sin\phi & 0 \\ \sin\phi & r\cos\phi & 0 \\ 0 & 0 & 1 \end{bmatrix}$$

But still where do I get the $$\phi$$? :\

Get the $$\phi$$ for what? It's okay to leave it as it is when you're in cylindrical coordinates.

 

[dingo_d]

Oh, so I need to find $$a_\rho$$, and that's just

$$a_\rho=2y\cos\phi-z\sin\phi$$ ?

Hmmm... I thought I had to calculate the phi :\

 

[planck42]

btw you should convert x's, y's, and z's to $$\rho$$'s, $$\phi$$'s, and z's/$$\theta$$'s

 

[dingo_d]

So then the solution is:

$$a_\rho=2\rho\sin\phi\cos\phi-z\sin\phi$$
$$a_\phi=-2\rho\sin^2\phi-z\cos\phi$$
$$a_z=3\rho\cos\phi$$

And to obtain $$\vec{a}$$ I just need to change unit vectors, and group them together to obtain the vector in cylindrical coordinate system?