Converting a Cartesian plane to spherical coordinates?

seralin

New member
Joined
Apr 9, 2019
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1
If I have the equation of a plane like z = 9 or y = 3, how can I rewrite them in spherical coordinates?
I know that with a point in 3D you would find ρ,θ,φ - for a plane like z = 9 how would I write ρ?
I'm guessing that θ might be 2π, but I'm lost as to how to find ρ and φ for a plane instead of a point.
Thanks!
 

Romsek

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Nov 16, 2013
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If you have some subset of \(\displaystyle \mathbb{R}^3\) specified by an equation in cartesian coordinates you can always
just substitute in the appropriate expressions in spherical coordinates.

\(\displaystyle x=\rho \sin(\theta)\cos(\phi),~y=\rho \sin(\theta)\sin(\phi),~z=\rho \sin(\theta)\)

for example \(\displaystyle z=9 \Rightarrow \rho \sin(\theta) = 9,~\rho = 9\csc(\theta)\)
 

Dr.Peterson

Elite Member
Joined
Nov 12, 2017
Messages
3,375
If I have the equation of a plane like z = 9 or y = 3, how can I rewrite them in spherical coordinates?
I know that with a point in 3D you would find ρ,θ,φ - for a plane like z = 9 how would I write ρ?
I'm guessing that θ might be 2π, but I'm lost as to how to find ρ and φ for a plane instead of a point.
Thanks!
I hope you understand that what you will get is not specific values of the coordinates (as for a point), but an equation in the three coordinates. Romsek gave you the full answer in the case of the plane z=9. In particular, θ does not have one value; it can have any value, and then ρ is a function of that. Moreover, φ is not present in his equation at all; it can have any value independent of the other coordinates.

You can do something similar for y=3, but then all three coordinates are involved.

But beware: not all sources define the coordinates in the same way, so it's possible that your θ and φ are swapped or otherwise modified from Romsek's version. Check for the conversion formulas as given in your own textbook!
 

Romsek

Full Member
Joined
Nov 16, 2013
Messages
251
If you have some subset of \(\displaystyle \mathbb{R}^3\) specified by an equation in cartesian coordinates you can always
just substitute in the appropriate expressions in spherical coordinates.

\(\displaystyle x=\rho \sin(\theta)\cos(\phi),~y=\rho \sin(\theta)\sin(\phi),~z=\rho \sin(\theta)\)

for example \(\displaystyle z=9 \Rightarrow \rho \sin(\theta) = 9,~\rho = 9\csc(\theta)\)
There is an error in above.

\(\displaystyle z = \rho \cos(\theta)\) and thus \(\displaystyle z=9 \Rightarrow \rho = 9 \sec(\theta)\)
my apologies
 
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