Asume I have an arbitrary, parametric, closed, orientable, surface; a sphere, ellipsoid, closed cylinder, weird general cone....

If you only have access to the parametrization, how can you determine whether a 3D point is inside or outside of the manifold?


1 Answer 1


I presume you are trying to determine is if a point $p$ is inside or outside the region $\Omega$ of space enclosed by the manifold, which can be seen as the boundary $\partial \Omega$ of $\Omega$. A solution is to use use Stokes–Cartan theorem: $$ \int_{\partial \Omega} \omega = \int_{\Omega} d\omega $$ with a suitable choice of the differential form $\omega$. In our case the following will work, compute: $$ I = \int_{\partial \Omega} \frac{(x-p)}{|x-p|^3} dS $$ over the manifold, where $x$ is the (3-dim) variable of integration, and $dS$ is the 2-form representing the area element of the surface. If $p$ is outside $\Omega$ then $I=0$. If $p$ is inside then $|I|$ yields the area of a unit sphere. Replace the exponent 3 in the denominator with the number of dimensions for a more general result, e.g., for $n=2$ (a curve in the 2-dim space) what you get is the winding-number of the curve around $p$ multiplied by $2\pi$.

  • 1
    $\begingroup$ I am going to stare really hard at this until i understand at least half $\endgroup$
    – Makogan
    Nov 29, 2018 at 23:00
  • $\begingroup$ Informally, think of the integrand as a small solid angle determined by rays traced from $p$ to a small portion of your surface around a point $x$ selected on the surface. E.g. see how it looks if your surface is a sphere of radius $r$ and $p$ is the origin of coordinates, then compute the integral in polar coord. On the other hand, if $p$ is outside the region enclosed by the manifold then by the Stokes theorem the integral has to be zero because our $\omega = \frac{(x-p)dS}{|x-p|^3}$ is a closed form and $d\omega = 0$. I believe I saw a similar example in Spivak's book on diff.geom. $\endgroup$
    – mlerma54
    Nov 30, 2018 at 3:34

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