It is way easier to think when a plane section of a cylinder, that is an ellipse having ratio between the axis that depends on the angle between the cutting plane and the axis of the cylinder, is also a plane section for a sphere i.e. a circle. There is no inaccuracy in point $(2)$: in order to have a planar intersection, the center of the sphere has to lie on the axis on the cylinder.
Center of circle: at $(0,0,3)$ , radius = $3$.
Plane $z = x +3$ passes through center.
Normal of plane: $(1,0,-1)$ .
Let vector $(a,b,c)$ be perpendicular to this normal:
$(a,b,c) \cdot (1,0,-1)$ = $0$ ; $a - c = 0$.
Yields 2 independent, orthogonal vectors perpendicular to the normal $(1,0,-1)$ of the plane:
1) $(1/√2)(1,0,1)$ and
2) $(0,1,0)$.
Let $\vec{s}$ = $\alpha (1/√2)(1,0,1) +\beta (0,1,0)$.
In analogy to a circle traced in the $x, y$ - plane:
$\vec{s} \cdot (1/√2)(1,0,1)$ = $3 cos(\theta)$ = $\alpha$.
$\vec{s} \cdot (0,1,0)$ = $3 sin(\theta)$ = $\beta$
Finally the parameter representation of the great circle:
$\vec{r}$ = $(0,0,3) + (1/√2)3cos(\theta)(1,0,1) + 3sin(\theta)(0,1,0)$
Best Answer
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
$$2\cos^{-1}\left(\frac{d^2+r_1^2-r_2^2}{2dr_1}\right)$$ 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
$$\frac{\pi}{12d}\left(d^4+8d(r_1^3+r_2^3)-3(r_1^2-r_2^2)^2-6d^2(r_1^2+r_2^2)\right)$$