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Given is the sphere $x^2 + y^2 + z^2 = 4$ and the plane $x + y = 2$ in $\mathbb R^3 $.

How can I find a parametrization for the intersection of the two?

user62608
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  • Begin by parametrizing the sphere. – ncmathsadist Feb 17 '13 at 01:18
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    @ncmathsadist It can't be done. Although the sphere is a smooth manifold, it is not a parametrisable manifold. – Fly by Night Feb 17 '13 at 01:46
  • You can write parametric equations for the sphere. Use sines and cosines. – ncmathsadist Feb 17 '13 at 02:01
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    @ncmathsadist It can't be done. You might be able to parametrise parts of the sphere, but you'll never find a regular parametrisation of the entire sphere. If you think otherwise, then why not supply such a parametrisation. If you could then you'd contradict results by Poincaré and Hopf. – Fly by Night Feb 17 '13 at 02:58
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    @ncmathsadist Perhaps you're thinking of $(\theta,\phi) \mapsto (\cos\theta\sin\phi,\sin\theta\sin\phi,\cos\phi)$? This does indeed cover the sphere, but it is not a regular parametrisation. The wheels come off when $\sin\phi=0$. This corresponds to the points $(0,0,\pm 1)$. The whole line $(\theta,0)$ goes to the point $(0,0,1)$ while the whole line $(\theta,\pi)$ goes to the point $(0,0,-1)$. – Fly by Night Feb 17 '13 at 03:12
  • Correct, but it could suffice for the purpose. – ncmathsadist Feb 17 '13 at 03:14
  • "Although the sphere is a smooth manifold, it is not a parametrisable manifold". Interesting. We bone-heads in the CG and CAD industries have been parameterising spheres for 40 years. What does "parametrisable" mean to a mathematician? Looks like it requires the map to be injective?? – bubba Feb 17 '13 at 05:23
  • @bubba It's the differential that needs to be injective, not the parametrisation itself. For example, the figure 8 is a parametrisable manifold even though it has a crossing. If we could find a regular parametrisation of a sphere by the plane then we would be saying that the sphere and the plane are somehow similar. I'm sure you can see that the plane is inherently different to the sphere.

    We'd need a parametrisation to cover the sphere and be an immersion at all points.

    Take a look at the following article:

    https://en.wikipedia.org/wiki/Immersion_%28mathematics%29

     – Fly by Night Feb 17 '13 at 16:18
  • OK. Thanks. I guess we are from different worlds. In my world, a parameterisation of a surface is just a mapping from a rectangle to $R^3$. We generally expect the mapping to be continuous, but not much more. – bubba Feb 18 '13 at 03:16
  • @bubba (Part 1) If you only want to cover the sphere then that's fine. You already know what the sphere looks like and that's fine. In differential geometry we want to use parametrisations to give information about the image. One way to study the singularities of a surface is to find a regular parametrisation and then look for the critical points and values of said parametrisation. If the parametrisation is not regular then it will introduce its own singularities. If we were to believe $(\theta,\phi) \mapsto (\cos\theta,\sin\theta\sin\phi,\cos\theta)$ then the sphere is not a smooth manifold! – Fly by Night Feb 18 '13 at 20:00
  • @bubba (Part 2) This very problem is why the definition of a manifold talks about local neighbourhoods. If every point has a neighbourhood that has a regular parametrisation (and these parametrisations overlap in a nice way) then it is a smooth manifold. To be a "parametrisable manifold" it needs a global, regular parametrisation. We can cover the sphere in charts. All of which are like little planes. – Fly by Night Feb 18 '13 at 20:01

4 Answers4

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You are in luck because one of the equations is linear.

Since $x+y=2$ it follows that $x=2-y$. We can substitute this into the first equation:

$$x^2+y^2+z^2=4 \implies (2-y)^2+y^2+z^2=4 \implies 2y^2-4y+z^2=0\,. $$

We can "complete the square" on the first part to give:

$$2[(y-1)^2-1]+z^2=0 \implies 2(y-1)^2+z^2=2 \, . $$

Considering only the $yz$-plane then we have an ellipse in the $yz$-plane parametrised by $(y(\theta),z(\theta)) = (1+\cos\theta,\sqrt{2}\sin\theta)$. We also know that $x=2-y$, hence:

$$\gamma(\theta) = (1-\cos\theta,1+\cos\theta,\sqrt{2}\sin\theta)$$

would give a regular parametrisation of the set in question.

Fly by Night
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Let $\mathbf c = (1,1,0)$, and define two orthogonal unit vectors by $\mathbf u = (1,-1,0)/\sqrt 2$, and $\mathbf v = (0,0,1).$ Then Fly By Night's equations can be written as:

$$\gamma(\theta) = \mathbf c + (\sqrt 2 \cos\theta)\mathbf u + (\sqrt 2 \sin\theta)\mathbf v $$

This makes it clear that the curve is a circle with center at $\mathbf c$ and radius $\sqrt 2$.

You only asked about the parametrisation of the curve, not its shape, so this is slightly off topic, I suppose, but it might still be of interest.

bubba
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The intersection is an ellipse:

$$x^2 + (2-x)^2 + z^2 = 4 \implies 2(x- 1)^2 +z^2 = 2$$

So parametrize as follows:

$$x=1 + \cos{t}$$ $$z= \sqrt{2} \sin{t}$$

Ron Gordon
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  • The intersection of a sphere and a plane is an ellipse?? That's true (since the curve is actually a circle, which is necessarily a special-case of an ellipse), but it's a bit misleading – bubba Feb 17 '13 at 05:15
  • The equation $2(x-1)^2 +z^2 = 2$ represents an elliptical cylinder whose axis runs in the $y$-direction. But the intersection of this cylinder with the given plane is actually a circle. – bubba Feb 17 '13 at 05:33
  • Ellipse in projection, a true circle in 3-space. An intersection of a sphere is always a circle. – Narasimham Mar 29 '15 at 14:12
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Try these equations.

$$\cases{ x=r \sin(s) \cos(t) \cr y=r \cos(s) \cos(t)\cr z=r \sin(t)\cr} $$

This is not a homeomorphism. The value $r$ is the radius of the sphere. It will parametrize the sphere for the right values of $s$ and $t$. This could be useful in parametrizing the ellipse. You can reparametrize if necessary to avoid any nastiness.

ncmathsadist
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  • This is messy because the "poles" of the sphere (as in Arctic and Antarctic) lies on the given plane. Better to put these poles on the $x$-axis or the $y$-axis, where they won't cause any trouble. Just swap the equations for $x$ and $z$, for example. – bubba Feb 17 '13 at 05:28
  • Yes, this could be shifted around to keep the poles off of the curve. – ncmathsadist Feb 17 '13 at 12:25