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Calculate divergence of <y^2,z^2,x^2> in cylindrical coords

  1. Oct 14, 2015 #1
    Hi everyone

    My professor just asked us a question that I can't get my head around. So we have the original vector in Cartesian format, <y^2,z^2,x^2>
    Then I am asked to convert to cylindrical coordinates:

    z= z;
    θ==arctan(z^2/y^2);
    r = \sqrt(y^4+z^4)

    However , I am then asked to take the divergence and curl of these items, I have the formula for both but I dont know how to fit them in!

    ∇⋅v=1r∂∂r(rvr)+1r∂vθ∂θ+∂vz∂z

    ∇×v=(1r∂vz∂θ−∂vθ∂z)r^+(∂vr∂z−∂vz∂r)θ^+(1r∂∂r(rvθ)−1r∂vr∂θ)z^

    For example, for div in tems of r, 1r∂∂r(r(sqrt(z^4+y^4))) which doesnt seem to work
     
  2. jcsd
  3. Oct 14, 2015 #2

    BvU

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    Hi tub,

    Can it be that your original vector is actually a vector function $$\ <x, y, z > \rightarrow <y^2, \ z^2,\ x^2 >$$ which you have to convert to a vector function in cylindrical coordinates $$\ <\rho, \phi, z > \rightarrow <f_1 (\rho, \phi,z) ,\ f_2 (\rho, \phi,z), \ f_3 (\rho, \phi,z) > \text ? $$
    The conversion (in my experience) goes like $$ \rho = \sqrt{x^2+y^2} \\ \phi = \arctan {y\over x} \quad ^{(*)} \\ z = z$$
    (*) with a caveat for ##\phi = \pi + \arctan {y\over x} \ \ \text {if} \ \ x< 0 \ ## to ensure ##\phi \in [-\pi/2, \ \pi/2]## or something similar
    and the reverse goes like $$x =\rho \cos\phi \\ y =\rho \sin \phi \\ z = z$$.​

    so that for f3 for instance, you get something like ##z\rightarrow \rho^2\cos^2\phi## ?

    once you have these fi you can let the operators go to work
     
  4. Oct 14, 2015 #3
    Ohhhhhh you have t change the variables inside r so that they are also in cyl coordinates than you so much! Is there an upvote button on this site?
     
  5. Oct 14, 2015 #4

    BvU

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    Appreciate your enthousiasm, but a) I have violated a PF rule by showing that example and b) you still have a lot of work to do !

    and c) I am not sure, it just seems a sensible perception of your exercise
     
  6. Oct 14, 2015 #5

    Mark44

    Staff: Mentor

    I don't have a problem with what you did.
     
  7. Oct 14, 2015 #6
    Ah Ok well thank you for your time!
     
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