Parametric representation of a plane cut of a sphere at y=5 The sphere is given by $x^2+y^2+z^2=36$
Parametric Form:
$$x=6\sin t\cos u$$
$$y=6\sin t\sin u$$
$$z=6\cos t$$
If the sphere is 'cut' at $z=5$ this problem is trivial. ($0<t<\arccos(5/6),0<u<2\pi$), but although the sphere cut at $y=5$ seems to be just as simple, using the standard angle definitions of the parametric form of a sphere, we now have 2 unknowns instead of 1. How do I go about this?

Choose for parametrization:
$$\begin{gathered}
  x = 6cos(t)cos(u) \hfill \\
  y = 6sin(t) \hfill \\
  z = 6cos(t)sin(u) \hfill \\ 
\end{gathered}$$
Rotate half circle around y-axis.
Then to solve
$$y = 6sin(t) = 5$$
is done as before. Choose right parametrization.
