Consider the formation of a lens by intersection of two spheres.
How can I calculate the cone angles formed for each spheres formed by the line connecting the centers of the spheres and the line passing through the center of a sphere and the circle formed by the intersection?
What happens when one sphere gets engulfed by the other (it should be total coverage or $2\pi$ radians for one, but what about the other?)?
let the radii of the spheres be $r_1$ & $r_2 $ & the distance between their centers be $d$ such that $$(r_1-r_2)<d<(r_1+r_2) \quad \text{where, } r_1\geq r_2$$ Now, assuming that one sphere with radius $r_1$ is centered at the origin (0, 0, 0) & another sphere with radius $r_2$ is centered at a point $(d, 0, 0)$ on the $x$ axis. then we can find out the distance of the common chord from the center of each sphere.
then the cone angle subtended by the plane (circle) of the intersection at the center of the sphere with radius $r_1$ is (refer to Intersection of two circles)
$$2\cos^{-1}\left(\frac{d^2+r_1^2-r_2^2}{2dr_1}\right)$$ and the cone angle subtended by the plane (circle) of the intersection at the center of the sphere with radius $r_2$ is
$$2\cos^{-1}\left(\frac{d^2+r_2^2-r_1^2}{2dr_2}\right)$$
In this case, the volume of intersection (engulfed) by the spheres is $$\frac{\pi}{3}\left(d^3+2r_1^3+2r_2^3-3Kr_1^2-3(d-K)(r_2^2+Kd)\right)$$ where, K is a constant & given as $$K=\frac{d^2+r_1^2-r_2^2}{2d}$$
by substituting the value of constant K & simplifying, the volume of the lens engulfed (common) by the intersecting spheres is
$$\frac{\pi}{12d}\left(d^4+8d(r_1^3+r_2^3)-3(r_1^2-r_2^2)^2-6d^2(r_1^2+r_2^2)\right)$$