Rectangular to Spherical Coordinate conversion....

In summary, the given point (-sqrt3/2, 3/2, 1) can be converted into spherical coordinates as (2, pi/3, 2pi/3), taking into account the possible variations in notation for theta and phi.
  • #1
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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  • #2
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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  • #3
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.
 
  • #4
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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  • #5
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...
 
  • #6
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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  • #7
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.
 
  • #8
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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  • #9
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.
 
  • #10
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.
 

Related to Rectangular to Spherical Coordinate conversion....

1. What is the purpose of converting from rectangular to spherical coordinates?

Converting from rectangular to spherical coordinates allows us to express a point in three-dimensional space using a different coordinate system. This can be useful in various applications, such as in physics and engineering, where spherical coordinates may be more intuitive or convenient to use.

2. How do you convert from rectangular to spherical coordinates?

To convert from rectangular to spherical coordinates, we use the following equations:

ρ = √(x² + y² + z²)

φ = arctan(y/x)

θ = arccos(z/ρ)

Where ρ is the distance from the origin, φ is the angle in the xy-plane, and θ is the angle from the positive z-axis.

3. What are the advantages of using spherical coordinates over rectangular coordinates?

Spherical coordinates can be advantageous in certain situations, such as when dealing with objects that have a spherical shape or when working with physical phenomena that involve spherical symmetry. They can also simplify certain calculations, such as finding the distance between two points or finding the volume of a sphere.

4. Can you convert from spherical to rectangular coordinates?

Yes, it is possible to convert from spherical to rectangular coordinates using the following equations:

x = ρsin(θ)cos(φ)

y = ρsin(θ)sin(φ)

z = ρcos(θ)

5. Are there any limitations to using spherical coordinates?

While spherical coordinates can be useful in certain situations, they do have some limitations. For example, they may not be as convenient to use when working with objects that have a non-spherical shape. Additionally, some calculations may be more complex in spherical coordinates compared to rectangular coordinates.

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