Parameterizing the boundary curve of the surface defined by $x+y+z \geq 1$ and $x^2 +y^2+z^2=1$

Question

I am unsure how to parameterize the boundary curve of the surface defined by

$x+y+z \geq 1$ and $x^2 +y^2+z^2=1$,

where $x,y$ and $z$ are real numbers.

The boundary curve should be the circle formed by the intersection of the plane $x+y+z = 1$ and sphere $x^2 +y^2+z^2=1$, but I am stuck when it comes to parameterizing this circle.

Further, is there a general strategy for parameterizing boundary curves of surfaces?

I would greatly appreciate any help!

(Aside: This question is in context of Stoke's theorem)

Answer 1

Since you already now that it is a circle, you need to

1. Find its center $C$
2. Find its radius $R$
3. Find an orthonormal basis $e_1, e_2$ of the plane this circle lies in

The first is easy: the center must be in the plane $x+y+z=1$, and by symmetry it should have all coordinates equal. Thus, $C = \left(\frac{1}{3},\frac{1}{3},\frac{1}{3}\right)$.

Next, the radius is also easy: just take the distance between $C$ and any point on the circle, say, $(1,0,0)$: $|C-(1,0,0)| = \sqrt{\frac{4}{9}+\frac{1}{9}+\frac{1}{9}}=\sqrt{\frac{2}{3}}$.

Finally, the plane $x+y+z=1$ is orthogonal to the vector $n=(1,1,1)$. There are numerous ways to find orthonormal $e1, e2$ orthogonal to $n$, e.q. one could take the basis $n, (1,0,0), (0,1,0)$ and apply Gram-Schmidt process to it. Alternatively, take the cross product of $n$ and $e_x$ (unit vector in $x$-direction): $n \times (1,0,0)= (0,1,-1)$, then normalize $e_1=\left(0, \frac{1}{\sqrt 2}, -\frac{1}{\sqrt 2}\right)$. Now, take the cross product again: $n \times e_1 = \left(-\sqrt 2, \frac{1}{\sqrt 2}, \frac{1}{\sqrt 2}\right)$ and normalize: $e_2 = \left(-\frac{\sqrt 2}{\sqrt 3}, \frac{1}{\sqrt 6}, \frac{1}{\sqrt 6}\right)$.

Finally, the circle is $C + R \cdot (e_1 \cos \theta + e_2 \sin \theta )$ for $\theta \in [0, 2 \pi)$.

Answer 2

The “boundary curve” is determined by $x+y+z=1$ and $x^2 + y^2 + z^2 = 1$.

You got 2 equations with 3 Unknown’s. You can solve this with one parameter. If you express x, z and z in terms of your parameter (let’s say you called it “t”). You can plug that back in one of the equations.

Answer 3

While the answer by Finn Eggers covers the general strategy you were after, the result produced by this strategy may fail to reveal the underlying symmetry (that of permutations between $x,y,z$) which was immediately apparent in the original equations.

To highlight the said symmetry, I propose the following parametrization: $$ \begin{align} x&={1\over3}+{2\over3}\cos(t)\\ y&={1\over3}+{2\over3}\cos\left(t+{2\pi\over3}\right) \\ z&={1\over3}+{2\over3}\cos\left(t+{4\pi\over3}\right) \end{align} $$