Vector Fields in Cartesian and Cylindrical Coordinates, The Curl

[sriracha]

All necessary information is attached except the answer in Cartesian coordinates, which is -ix-jy+2kz and my work converting back from cylindrical to Cartesian, which I used WolframAlpha for, as the trig is a mess (that is, if the way I am doing this is correct).

http://www.wolframalpha.com/input/?i=%28icos%28tan^-1%28y%2Fx%29%29%2Bjsin%28tan^-1%28y%2Fx%29%29%29%28-sqrt%28x^2%2By^2%29%28cos^2%28tan^-1%28y%2Fx%29%29-sin^2%28tan^-1%28y%2Fx%29%29%29%29%2B%28-isin%28tan^-1%28y%2Fx%29%29%2Bjcos%28tan^-1%28y%2Fx%29%29%29%282z%28cos^2%28tan^-1%28y%2Fx%29%29-sin^2%28tan^-1%28y%2Fx%29%29%29%29&a=i_Variable

WolframAlpha, however, just gave me this mess back. Would appreciate if someone would let me know if the way I'm going about this is correct. I have a strong feeling the matrix I'm using for reconversion is off.

Edit: I already posted this in the physics section, but I got no replies, because it is probably more appropriate to place here. See the attachment on the original thread at https://www.physicsforums.com/showthread.php?t=588902

 

[lippyka]

for the full problem.The answer in Cartesian coordinates is -ix-jy+2kz.To convert from cylindrical to Cartesian coordinates, you can use the following transformation matrix: \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} \cos(\theta) & -\sin(\theta) & 0 \\ \sin(\theta) & \cos(\theta) & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} r \\ \theta \\ z \end{bmatrix}.Using this transformation matrix, you can obtain the Cartesian coordinates of the point in question by plugging in the given cylindrical coordinates:\begin{bmatrix} x \\ y \\ z \end{bmatrix}= \begin{bmatrix} \cos(\tan^{-1}(y/x)) & -\sin(\tan^{-1}(y/x)) & 0 \\ \sin(\tan^{-1}(y/x)) & \cos(\tan^{-1}(y/x)) & 0 \\ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} \sqrt{x^2 + y^2} \\ \tan^{-1}(y/x) \\ 2z \end{bmatrix}= \begin{bmatrix} -x \\ -y \\ 2z \end{bmatrix}.Therefore, the answer in Cartesian coordinates is -ix-jy+2kz.