Converting a Cartesian plane to spherical coordinates?

seralin

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Apr 9, 2019
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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!
 
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 \(\displaystyle z=9 \Rightarrow \rho \sin(\theta) = 9,~\rho = 9\csc(\theta)\)
 
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!
 
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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