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

[tub08918]

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 don't 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 doesn't seem to work

 

[BvU]

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 f_i you can let the operators go to work

 

[tub08918]

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?

 

[BvU]

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

 

[Mark44]

Appreciate your enthousiasm, but a) I have violated a PF rule by showing that example

I don't have a problem with what you did.

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

 

[tub08918]

Ah Ok well thank you for your time!