Rectangular to Spherical Coordinate conversion....

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The discussion revolves around converting rectangular coordinates (-(sqrt3)/2, 3/2, 1) to spherical coordinates. The participant initially calculated the spherical coordinates as (2, -pi/3, pi/3) but faced confusion regarding the signs and formulas used. It was clarified that the sign of x affects the calculation of theta, leading to potential errors in the result. Different conventions for spherical coordinates in physics and mathematics were discussed, highlighting variations in definitions for angles. Ultimately, the participant resolved their confusion and arrived at the correct answer, acknowledging the importance of understanding these conventions.
Unicow
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Homework Statement


Convert from rectangular to spherical coordinates.
(-(sqrt3)/2 , 3/2 , 1)

Homework Equations


We know the given equations are
ρ = sqrt(x^2 + y^2 + z^2)
tan theta = y/x
cos φ = z / ρ

The Attempt at a Solution


My answer was (2, -pi/3, pi/3)
It should be a simple plug and go... Am I doing simple math wrong? The only part I think I could get wrong is the y/x but -3 / 2 / sqrt3 / 2 should be sqrt3 right?I'm using webassign if that matters at all. I'm sorry for posting such a simple question but I don't understand how that solution is wrong... I guess I'm just having a huge brain fart?
 
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Unicow said:

Homework Statement


Convert from rectangular to spherical coordinates.
(-(sqrt3)/2 , 3/2 , 1)

Homework Equations


We know the given equations are
ρ = sqrt(x^2 + y^2 + z^2)
tan theta = y/x
cos φ = z / ρ

The Attempt at a Solution


My answer was (2, -pi/3, pi/3)
It should be a simple plug and go... Am I doing simple math wrong? The only part I think I could get wrong is the y/x but 3 / 2 / sqrt3 / 2 should be sqrt3 right?
No, since ##x = \frac{-\sqrt 3}{2}##. You are ignoring the sign, so your value for ##\theta## will be wrong.
Unicow said:
I'm using webassign if that matters at all. I'm sorry for posting such a simple question but I don't understand how that solution is wrong... I guess I'm just having a huge brain fart?
 
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Mark44 said:
No, since ##x = \frac{-\sqrt 3}{2}##. You are ignoring the sign, so your value for ##\theta## will be wrong.

Sorry I must have deleted the negative sign by accident while I was editing. I've edited it back in and I had counted that into my calculations and that's why my theta is negative pi / 3 instead of positive.
 
Your answer of (2, -pi/3, pi/3) looks fine to me. Is it possible that WebAssign doesn't recognize "pi" and wants you to enter a decimal value?
Sometimes these programs are brain-dead...
 
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Mark44 said:
Your answer of (2, -pi/3, pi/3) looks fine to me. Is it possible that WebAssign doesn't recognize "pi" and wants you to enter a decimal value?
Sometimes these programs are brain-dead...

Yeah it's really frustrating especially considering it's such an easy question. I've tried decimal already and the simple "pi" or whatever, is correct for all the other choices...
 
Unicow said:

Homework Equations


We know the given equations are
ρ = sqrt(x^2 + y^2 + z^2)
tan theta = y/x
cos φ = z / ρ
The second and third equations above are wrong, according to this wiki article (https://en.wikipedia.org/wiki/Spherical_coordinate_system).
##\theta = \arccos(z/r)## Here r is the same as your ##\rho##.
##\phi = \arctan(y/x)##
 
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Mark44 said:
The second and third equations above are wrong, according to this wiki article (https://en.wikipedia.org/wiki/Spherical_coordinate_system).
##\theta = \arccos(z/r)## Here r is the same as your ##\rho##.
##\phi = \arctan(y/x)##

There's no way that can be right... I've already done a few questions and gotten them all right except for this one... Also, I might trust the textbook more than wikipedia. I hope you don't take this as an insult or anything because I do appreciate the help.
 
Unicow said:
There's no way that can be right... I've already done a few questions and gotten them all right except for this one... Also, I might trust the textbook more than wikipedia. I hope you don't take this as an insult or anything because I do appreciate the help.
No, I don't take it as an insult at all. The wiki formulas seem weird to me, as well, and that's why I qualified my answer.

I checked with wolframalpha, which gives an answer of (2, pi/3, 2pi/3) -- http://www.wolframalpha.com/input/?i=spherical+coordinates(-sqrt(3)/2,+3/2,+1).
Not all books use the notation in the same way. For a point ##(\rho, \theta, \phi)##, some books call ##\theta## the inclination (measured away from the z-axis), and others call ##\phi## the inclination. These differences make your formulas correct as far as some books are concerned and incorrect in others.
 
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Mark44 said:
No, I don't take it as an insult at all. The wiki formulas seem weird to me, as well, and that's why I qualified my answer.

I checked with wolframalpha, which gives an answer of (2, pi/3, 2pi/3) -- http://www.wolframalpha.com/input/?i=spherical+coordinates(-sqrt(3)/2,+3/2,+1).
Not all books use the notation in the same way. For a point ##(\rho, \theta, \phi)##, some books call ##\theta## the inclination (measured away from the z-axis), and others call ##\phi## the inclination. These differences make your formulas correct as far as some books are concerned and incorrect in others.

Ahh I see and understand now haha. Thank you for your help and it completely went over my head about how it could be more than one value of "n * pi / 3" for theta to be the value. I got the correct answer! Thank you so so much.
 
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The typical convention in physics is that ##\theta## is the polar angle and ##\varphi## the azimuthal angle. In mathematics, the convention is usually the other way around. The Wikipedia page shows both conventions.
 

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