How SphericalPlot3D actually works?

Questions
,
1

I must do a graphical representation of a surface which is expressed in spherical coordinates. For this, I saw that Mathematica has a straightforward method with the SphericalPlot3D function.
Having x_1, x_2 and x_3 in my code as the components of the vector:

I^2=x_1^2+x_2^2+x_3^2

I just wrote the function and then applied the SphericalPlot3D method.

(*Constant values*)
A1 = 1;
A2 = 3;
A3 = 6;
SPIN = 15;
(*components of my vector*)
x1[spin_, theta_] := spin*Cos[theta];
x2[spin_, theta_, fi_] := spin*Sin[theta] Cos[fi];
x3[spin_, theta_, fi_] := spin*Sin[theta] Sin[fi];

(*angular momentum I squared*)
Spin2[spin_, theta_, fi_] := 
  x1[spin, theta]^2 + x2[spin, theta, fi]^2 + x3[spin, theta, fi]^2;

s1 = SphericalPlot3D[
  Sqrt[Spin2[SPIN, x, y]], {x, 0, \[Pi]}, {y, 0, 2 \[Pi]}, 
  LabelStyle -> {14, Bold, Black, FontFamily -> "Times New Roman"}]
Show[s1]

However, the plot which I obtain has values on the 3 axes which are confusing for me: I expect that on two of the three axes (e.g. the one that corresponds to polar angle \theta and azimuthal angle \varphi) to have values no more than 3.14 and 6.28, but al three axes have values which are larger than those.

Is my understanding of a spherical plot in Mathematica wrong?

Thank you in advance!

2

You are plotting the value of a constant function on the sphere:

In[23]:= Sqrt[Spin2[SPIN, x, y]] // Simplify

Out[23]= 15

This way you get a sphere with radius 15. Your coordinates x,y are not cartesian: they fix a direction in space.

Compare what you get with PolarPlot[15,{theta,0,2Pi}]: this associates with every direction theta a radius with length 15, so that you get a circle with radius 15.