Circle to cylindrical coordinates

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  • #1
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Homework Statement


Transform to cylindrical coordinates:
x[tex]^{2}[/tex]+y[tex]^{2}[/tex]=R[tex]^{2}[/tex]

Doesn't look like a problem at all first... :smile:

Homework Equations


.. after all I know that is a circle (2d) and we can forget the z-axis (=0) and transform it to just polar coords.
Also I know, that for polar coordinates
x = r cos[tex]\theta[/tex]
y = r sin[tex]\theta[/tex]

The Attempt at a Solution


But I don't know how to put it out as an answer. Am I supposed to give a parametric equation in terms of x=... and y=... or should it be like r=... and [tex]\theta[/tex]=... and z=0 or can I express it without parametres? :confused:
 

Answers and Replies

  • #2
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:surprised
hey, do I get the correct equation just by replacing x with r cos[tex]\theta[/tex] and y with r sin[tex]\theta[/tex]? So that cartesian eq
x[tex]^{2}[/tex]+y[tex]^{2}[/tex]=R[tex]^{2}[/tex]
is
(r cos[tex]\theta[/tex])[tex]^{2}[/tex]+(r sin[tex]\theta[/tex])[tex]^{2}[/tex]=R[tex]^{2}[/tex]
in cylindrical?

Or should it be something else, like parametric form of some kind?
 
  • #3
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:bugeye:
(r cos[tex]\theta[/tex])[tex]^{2}[/tex]+(r sin[tex]\theta[/tex])[tex]^{2}[/tex]=R[tex]^{2}[/tex]
and it all goes down to r=R?

I guess it goes the same way with the spherical coordinates.

I'm sorry for this monoloque. :blushing:

I hope someone would correct me if I'm wrong. thank you.
 
  • #4
HallsofIvy
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Yes, [itex]x^2+ y^2= R^2[/itex] is a cylinder (not a circle) of radius R so it is particulary simple in cylindrical coordinates.

No, it does not "go that way" with spherical coordinates. In spherical coordinates [itex]x= \rho cos(\theta) sin(\phi)[/itex] and [itex]\phi= \rho sin(\theta) sin(\phi)[/itex] so [itex]x^2+ y^2= \rho^2 cos^2(\theta)sin^2(\phi)+ \rho^2 sin^2(\theta)sin^2(\phi)[/itex][itex]=
\rho^2 sin^2(\phi)= R^2[/itex].
 
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  • #5
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Yes, [itex]x^2+ y^2= R^2[/itex] is a cylinder (not a circle) of radius R so it is particulary simple in cylindrical coordinates.
I think (and still do) that it is a circle, because it has no z. :eek:

No, it does not "go that way" with spherical coordinates. In spherical coordinates [itex]x= \rho cos(\theta) sin(\phi)[/itex] and [itex]\phi= \rho sin(\theta) sin(\phi)[/itex] so [itex]x^2+ y^2= \rho^2 cos^2(\theta)sin^2(\phi)+ rho^2 sin^2(\theta)sin^2(\phi)[/itex][itex]=
\rho^2 sin^2(\phi)= R^2[/itex].
Isn't the equation for circle R=r in both cylindrical and spherical coordinates?
 
  • #6
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I think (and still do) that it is a circle, because it has no z. :eek:



Isn't the equation for circle R=r in both cylindrical and spherical coordinates?
It is a cylinder, as a matter of fact. Because, it is saying that for any z at all, we have x^2+y^2=R^2.:tongue:
 
  • #7
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It is a cylinder, as a matter of fact. Because, it is saying that for any z at all, we have x^2+y^2=R^2.:tongue:
CylindricalCoordinates_1001.gif


If the z is zero, aren't we only left with a circle?

man.. understanding one thing requires not understanding two.. :yuck:

I hope I know how to turn cartesian equations to cylindrical and spherical now.

I'm really trusting on these tricks:

For cylindrical: x^2+y^2=r^2
For spherical: x^2+y^2+z^2=[tex]\rho[/tex]^2

So if I must for example turn this cartesian
x^2+y^2-z=0
into a speherical equation, i just add both sides z^2 and
x^2+y^2+z^2=z+z^2
[tex]\rho[/tex]^2=z+z^2
[tex]\rho[/tex]=[tex]\sqrt{z+z^2}[/tex] and because z= [tex]\rho[/tex] cos [tex]\phi[/tex]
[tex]\rho[/tex]=[tex]\sqrt{\rho cos \phi +(\rho cos \phi)^2}[/tex]

Most likely I'm wrong. Once again, help is really appreciated. :frown:
 

Attachments

  • #8
1,631
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CylindricalCoordinates_1001.gif


If the z is zero, aren't we only left with a circle?

man.. understanding one thing requires not understanding two.. :yuck:

I hope I know how to turn cartesian equations to cylindrical and spherical now.

I'm really trusting on these tricks:

For cylindrical: x^2+y^2=r^2
For spherical: x^2+y^2+z^2=[tex]\rho[/tex]^2

So if I must for example turn this cartesian
x^2+y^2-z=0
into a speherical equation, i just add both sides z^2 and
x^2+y^2+z^2=z+z^2
[tex]\rho[/tex]^2=z+z^2
[tex]\rho[/tex]=[tex]\sqrt{z+z^2}[/tex] and because z= [tex]\rho[/tex] cos [tex]\phi[/tex]
[tex]\rho[/tex]=[tex]\sqrt{\rho cos \phi +(\rho cos \phi)^2}[/tex]

Most likely I'm wrong. Once again, help is really appreciated. :frown:

That is one way of doing it. You could have also just substitute the values of x,y and z in terms of ro, cos, sin and corresponding angles, and you would have ended up with the same thing.

Note: the reason you can take the square root of [tex]\rho^2[/tex] is because it cannot be negative. Otherwise, you would have stepped on the red line...:biggrin:
 
  • #9
1,631
4
CylindricalCoordinates_1001.gif


If the z is zero, aren't we only left with a circle?
Well, as a matter of fact, the levle curves of this surface projected in any plane z=k, where k is a constant, are circles.
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
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956
CylindricalCoordinates_1001.gif


If the z is zero, aren't we only left with a circle?
Yes, but you are NOT told that z= 0. Because "it has no z", there is no restriction on z: z can be any number.

man.. understanding one thing requires not understanding two.. :yuck:[\quote]
No, understanding one thing requires understanding two!

I hope I know how to turn cartesian equations to cylindrical and spherical now.

I'm really trusting on these tricks:

For cylindrical: x^2+y^2=r^2
For spherical: x^2+y^2+z^2=[tex]\rho[/tex]^2

So if I must for example turn this cartesian
x^2+y^2-z=0
into a speherical equation, i just add both sides z^2 and
x^2+y^2+z^2=z+z^2
[tex]\rho[/tex]^2=z+z^2
[tex]\rho[/tex]=[tex]\sqrt{z+z^2}[/tex] and because z= [tex]\rho[/tex] cos [tex]\phi[/tex]
[tex]\rho[/tex]=[tex]\sqrt{\rho cos \phi +(\rho cos \phi)^2}[/tex]
It would be better to leave it as
[tex]\rho^2[/tex]=[tex]\rho cos \phi +(\rho cos \phi)^2[/tex]


Most likely I'm wrong. Once again, help is really appreciated. :frown:
 

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