So I have a point $(r, \phi, z)$ which I can express in the cartesian coordinate system as $(r cos \phi, r sin \phi, z)$. If I would convert the components of the vector (to cylindrical coordinates) $ \begin{bmatrix} r cos \phi \\ r sin \phi \\ z \end{bmatrix} $ by multiplying with transformation matrix $ \begin{bmatrix} cos \phi & sin \phi & 0 \\ -sin \phi & cos \phi & 0 \\ 0 & 0 & 1 \end{bmatrix} $ we get $ \begin{bmatrix} r \\ 0 \\ z \end{bmatrix} $. My question is why the $\phi$ component of result zero ? What does it ($\phi$ component) signify here in case of a point on a surface of cylinder ? I thought about this, and I think since the cylindrical coordinate system was made up by a radius and the rotation of point, direction of $\phi$ doesn't have any sense here hence it is zero here. Is this correct ? If so, is there any mathematical definition for this ?
Edit: I posted this question because I was trying to prove divergence theorem on a closed cylindrical surface ($\pi$/4 < $\phi$ < $\pi$/2; 0 < z < 1/2; $r=2$), I took a differential area on the side rectangles $drdz$, but could not figure direction of area in terms of vector component. Should it be $\hat{\phi}$ ($-\hat{\phi}$ for $\pi/4$) or 0 as we found for transformation matrix ? My intuition says it would be $\hat{\phi}$ but I can't explain the 0.
In 3D space we may draw a line through the origin, and consider every 2D plane which contains that line. Every point $P$ in the space which lies away from the line belongs to exactly one such plane. To describe $P$ is cylindrical coordinates, we first give its 2D rectangular coordinates $r$ and $z$ in the plane which contains it. Then we specify the plane using an angle $\theta$.