Consider the unit 2-sphere $\mathbb{S}^2$. Let there be an infinite cone with its vertex at the center of the sphere. Let the symmetry axis of the cone intersect $\mathbb{S}^2$ at the point P within the cone. My question regards an infinitesimal neighborhood of $P$ lying entirely within $\mathbb{S}^2$. I want to know if this neighborhood is a flat disc or not. Specifically, I want to know if I can properly establish a system of plane polar coordinates to describe points in the patch of the form $(r,\theta)$ with $r=0$ at $P$. Obviously $\theta\in[0,2\pi]$ and $r$ would be some fraction of the infinitesimal radius. I don't want to use the spherical polar coordinates to analyze the patch. I want to use the plane polar coordinates.
The Gaussian curvature $K$ is uniformly $1/r^2=1$ everywhere on the unit sphere. If, in general, the Gaussian curvature is the ratio of the solid angle subtended by the normal projection of a small patch divided by the area of that patch, then
$$ K = \frac{\Omega}{A} ~~.$$
If the cone's opening angle is $2\theta$, then the formula for the solid angle of the section of the sphere within the cone is
$$ \Omega=2\pi(1-cos\theta) ~~. $$
From Archimedes we know that the area of the spherical cap of the cone is
$$ A = 2\pi R h $$
where $R$ is the radius of the sphere and $h$ is the height of the cap. Let $\delta z$ be the infinitesimal height of the spherical cap formed by the infintiesimal neighborhood of $P$, and let $\delta\theta$ be the infinitesimal opening half-angle of the cone. Then
$$ K = \dfrac{2\pi(1-cos\delta\theta) }{2\pi\delta z} = \dfrac{ 1-cos\delta\theta }{ \delta z} = \dfrac{1-(1-\delta\theta^2/2+...)}{\delta z} $$
If we take the usual approach that the product of infinitesimals is zero then
$$ K = \dfrac{0}{\delta z}=0 ~~,$$
but actually the quadratic term in the cosine expansion looks like
$$ K := \dfrac{\delta\theta^2}{\delta z} $$
which is of the order of an infinitesimal and should not disappear. So... I don't know... can I treat the spherical cap as perfectly flat when it the cap is only an infinitesimal neighborhood of the point $P$? Thanks!
The precise answer is: there is no such thing as an infinitesimal piece of surface. The concept of infinitesimal objects is a point of view adopted by physicists because it helps them think, somehow, but it has no mathematical existence. You would have to define them entirely, and then define a concept of curvature for such objects.
A more tolerant answer is: a standard sphere has constant curvature, and the only reasonable meaning to "the curvature of an infinitesimal patch" is what you integrate over the sphere when you compute its global curvature. Since it is constant over the whole sphere, and since you get something non-zero when you integrate that constant, it must mean that the constant is non-zero.