# How can you determine if a point is inside a parametric 2D manifold?

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?

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 $$2\pi$$.
• 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. Nov 30, 2018 at 3:34