Parametric representation of a plane cut of a sphere at y=5

Question

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?

Answer 1

Introduce a new angle in the plane of intersection as $v$.

$ t =\cos^{-1}(5/6),$ a fixed latitude t for all $ 0<u<2\pi. $

Cut radius $\rho= \sqrt{6^2 -5^2} = \sqrt{ 11} $

$ (x = \rho \cos v, y = \rho \sin v, z = 5) $

Answer 2

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.

Answer 3

I think what you want is what values of $(t,u)$ belong to the plane $y=5$. Since that intersection of a sphere and a plane is a circle (2D curve) the result will be an implicit equation in terms of $t$ and $u$. In fact, you have

$$\sin(t) \sin(u) - \frac{5}{6}=0$$

The result is a closed curve wrapping around $(t=\frac{\pi}{2},u=\frac{\pi}{2})$. So your question is now how do you parametrize this implicit curve such that $t=t(s)$ and $u=u(s)$?

One way is to make $$ t(s) = \arcsin\left(s \sqrt\frac{5}{6}\right) \\u(s) = \arcsin\left(\frac{1}{s} \sqrt\frac{5}{6}\right)$$

for $s=\sqrt{\frac{5}{6}} \ldots \sqrt{\frac{6}{5}}$

In general, if the normal direction of the plane has a unit vector $(n_x,n_y,n_z)$ and a distance $d$ from the origin, then the implicit equation of the cut curve is

$$ n_z \cos(t)+n_x \sin(t)\cos(u)+n_y \sin(t)\sin(u) - \frac{d}{r} =0 $$