Vector field in cylindrical coordinates

[Les talons]

Homework Statement

Sketch each of the following vector fields.
[itex] E_5 = \hat \phi r [/itex]
[itex] E_6 = \hat r \sin(\phi) [/itex]

I wish to determine the [itex] \hat x [/itex] and [itex] \hat y [/itex] components for the vector fields so that I can plot them using the quiver function in MATLAB.

Homework Equations

A cylindrical coordinate is given by [itex] (r, \phi, z) [/itex]
[itex] r = \sqrt[+]{x^2 + y^2} [/itex]
[itex] \phi = \arctan(y/x) [/itex]
[itex] z = z [/itex]
Transformation from cylindrical coordinates into Cartesian coordinates:
[itex] \hat x = \hat r \cos(\phi) - \hat \phi \sin(\phi) [/itex]
[itex] \hat y = \hat r \sin(\phi) + \hat \phi \cos(\phi) [/itex]

The Attempt at a Solution

[itex] E_5 = \hat \phi r [/itex]
Since the given vector field [itex] E_5 [/itex] only has a [itex] \hat \phi [/itex] component,
[itex] \hat x = \hat r \cos(\phi) - \hat \phi \sin(\phi) = 0 \cos(\phi) - r \sin(\phi) = - \sqrt[+]{x^2 + y^2} \sin(\phi) [/itex]
Then I transformed [itex] \phi [/itex] in terms of [itex] x [/itex] and [itex] y [/itex] based on the definition of a cylindrical coordinate
[itex] \hat x = - \sqrt{x^2 + y^2} \sin(\arctan(y/x)) = \displaystyle \frac{-y \sqrt{x^2 + y^2} }{x \sqrt{1 +y^2 /x^2} } [/itex]
I followed the same approach for the [itex] \hat y [/itex] component
[itex] \hat y = \hat r \sin(\phi) + \hat \phi \cos(\phi) = 0 \sin(\phi) + r \cos(\phi) = \sqrt[+]{x^2 + y^2} \cos(\phi) [/itex]
[itex] \hat y = \sqrt{x^2 + y^2} \cos(\arctan(y/x)) = \displaystyle \frac{\sqrt{x^2 + y^2} }{\sqrt{1 +y^2 /x^2} } [/itex]
Then
[itex] E_5 = - \hat x \displaystyle \frac{y \sqrt{x^2 + y^2} }{x \sqrt{1 +y^2 /x^2} } + \hat y \displaystyle \frac{\sqrt{x^2 + y^2} }{\sqrt{1 +y^2 /x^2} }[/itex]

Is this approach valid? Thanks for helping. I'll post my attempt for [itex] E_6 [/itex] in a little while.

I'm having trouble thinking about the vector field since I'm used to working in Cartesian coordinates, so I can't tell if this makes sense.

 

[Les talons]

[itex] E_6 = \hat r \sin(\phi) [/itex]
[itex] \hat x = \sin(\phi) \cos(\phi) = \sin(\arctan(y/x)) \cos(\arctan(y/x)) [/itex]
[itex] \hat y = \sin^2(\phi) = \sin^2(\arctan(y/x)) [/itex]

 

[Les talons]

If anyone happens upon this thread looking for help in a similar question, the correct conversions are as follows for the [itex] \hat x [/itex] and [itex] \hat y [/itex] components of [itex] \vec E_5 [/itex]:
[itex] E_{5x} = E_{5r} \cos \phi - E_{5\phi} \sin \phi [/itex]
[itex] E_{5y} = E_{5r} \sin \phi + E_{5\phi} \cos \phi [/itex]

and similarly for [itex] \vec E_6 [/itex] (or another vector in cylindrical coordinates)