A small math question today.

Given a unit sphere and an arbitrary ellipsoid, calculate the surface area of sphere inside the ellipsoid.

Without loss of generality, we may assume the unit sphere to be centered at origin, then the question depends solely on the 6 parameters defining the ellipsoid, $c_i, \alpha_i$ for $i=1,2,3$ in $\sum (x_i-c_i)^2/\alpha _i ^2 = 1$. Without thinking too much about existence of analytic solution in terms of the parameters, we want to look for ways to calculate or at least approximate the problem.

My own solution is based on the famous Gauss-Bonnet theorem:

$\int _M KdA + \int _{\partial M} \kappa _g ds = 2\pi \chi (M)$

The curvature of a unit sphere is constantly 1, and the Euler characteristic of a disc is 1. As long as you can calculate the geodesic curvature you get

$|M| = 2\pi - \int _{\partial M} \kappa _g ds$.

The intersection is a union of finitely many closed curves we can just calculate one by one.

A simple example would be as follows:

Consider the intersection between $x^2+y^2+z^2=1$ and $(x-1)^2+y^2+z^2=1$. The intersection $\Gamma$ is given by $x = \frac{1}{2}$ and $y^2+z^2 = \frac{3}{4}$. This is a circle of radius $\frac{\sqrt{3}}{2}$ with geodesic curvature $\frac{1}{\sqrt{3}}$. Therefore we have $\int _{\Gamma} \kappa _g ds = \pi$, so $|A| = 2\pi \pm \pi$ depending on the orientation. The surface area can be verified using the formula from Archimedes $|A| = \pi (h^2+a^2)$.

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Of course, obtaining the parametrization is sometimes too difficult. Are there any easier ways to approximate the answer? Here are two suggestions from my friends.

- Monte Carlo based: sample points on the surface of the sphere and check whether it is in the ellipsoid.

- Integral approximation: similar to Monte Carlo but instead we split the sphere into pieces using spherical coordinates, then the required area is the integral over the indicator function of the ellipsoid.

And of course we can get an estimation using the Gauss-Bonnet approach. Let us call the sphere $S$ and the ellipsoid $E$. We start from a point of intersection, say $x$. The curve goes in the direction $v\in T_xS \cap T_xE$, the intersection of the tangent spaces of the two objects at $x$. We know the intersection must be of dimension 1 because if the two objects are cotangent at the point then the intersection is a point which generates no area after all.

Obtaining this direction is simply algebra, where we find the intersection between two subspace. With this direction we can numerically obtain a closed curve. There is a gap here because we may not go back to the starting point using numerical approximation. We just assume by smoothness (well sphere and ellipsoid are very nice objects) by doing fine enough approximation things would work (actually I believe we can prove this but I am not an expert in numerical methods) -- then we have an approximated curve of intersection.

From here we can calculate its length as well as its geodesic curvature, and the estimated area follows.

Gauss-Bonnet seems quite universal even if we generalize sphere and ellipsoid to other bounded closed $C^{\infty}$ objects because they are manifolds anyway. The only worry being whether we can get a closed curve without any global features that promises convergence.

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Well this is just a funny little problem that I enjoyed during the lockdown. I feel like I should write some maths before I go back to my creation whether it is literature, music or research.

Meanwhile I hope everyone is alright during the pandemic -- whether you call it COVID-19, or Wuhan virus. Stay indoor, and stay healthy.

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