Ex. Convert the following vector from Cartesian to Spherical
Coordinates and verify that its magnitude is the same in both
systems.
r  3xˆ  4yˆ  5zˆ
Recall that the representation of a point P in a particular system is given by just
listing the 3 corresponding coordinates in triplet form:
 x, y , z 
Cartesian
 r , , 
Spherical
and that we could convert the point P’s location from one coordinate system to
another using coordinate transformations.
Cartesian  Spherical
Spherical  Cartesian
r  x2  y2  z 2
 x2  y 2
  tan 

z

 y
  tan 1  
x
1
x  r sin  cos 
y  r sin  sin 
z  r cos




To convert the vector from Cartesian to Spherical Coordinates, we must convert
the xˆ , yˆ , zˆ unit vectors in to rˆ , θˆ , φˆ unit vectors using:
xˆ  sin  cos  rˆ  cos cos  θˆ  sin  φˆ
yˆ  sin  sin  rˆ  cos sin  θˆ  cos  φˆ
zˆ  cos rˆ  sin  θˆ
The expression that needs to be evaluated looks like this:

r  3 sin  cos  rˆ  cos cos  θˆ  sin φˆ


 4 sin  sin  rˆ  cos sin θˆ  cos  φˆ

 5 cos rˆ  sin  θˆ


Before r can be determined,  and must be found.
 x2  y 2
  tan 

z

1
 y
 

 32  42
1
  tan 


5


 4
 
  tan 1    tan 1    .9272952
x
3
 

 4

(45o )
(53.13o )
Rearranging r to group like terms together yields:
r   3sin  cos   4sin  sin   5cos  rˆ
  3cos cos   4cos sin   5sin   θˆ
  3sin   4cos   φˆ
Substituting our values for  and , we get:

 
 
  
r   3sin   cos .9272952   4sin   sin .9272952   5cos    rˆ
4
4
 4 


 
 
  
  3cos   cos .9272952   4cos   sin .9272952   5sin    θˆ
4
4
 4 

  3sin .9272952   4cos .9272952   φˆ
Reducing yields:
  2
 2
 2 


.6
4
.8
5
r   3








  rˆ

2
2
2





 
  2
 2
 2  ˆ
  3
.6   4 
.8   5 




  θ

2
2
2







  3 .8   4 .6   φˆ
 
r   5 2  rˆ
r  5 2 rˆ   0  θˆ 
 0  φˆ
Therefore, the equivalent (but not unique) vector in spherical coordinates is:


r  5 2 rˆ

r will only be unique if  and are given.
Otherwise, r
sphere.
would map out the surface of a
Magnitudes:
In Cartesian coordinates, the magnitude of r is given by:
r  x2  y2  z 2

r  32  42  52  50  5 2
In Spherical coordinates, the magnitude of r is given by:
r r

r 5 2