How can I get the volume enclosed by two orthogonal cylindrical surfaces?

Question

With the following code, I can only get the results when the radius of the ball and the half apex angle of the cone are specific values.

Clear["Global`*"];
v1 = ImplicitRegion[x^2 + y^2 + (z - 2)^2 <= 2^2, {x, y, z}]
v2 = ImplicitRegion[z >= Cot[Pi/3]*(\!\(TraditionalForm\`
\*SqrtBox[\(
\*SuperscriptBox[\(x\), \(2\)] + 
\*SuperscriptBox[\(y\), \(2\)]\)]\)), {x, y, z}]
v = RegionIntersection[v1, v2];
Assuming[a > 0 && Pi/2 > \!\(TraditionalForm\`\[Alpha]\) > 0, 
 Volume[v]]

(*10Pi*)

But very long time by:

Clear["Global`*"];
v1 = Ball[{0, 0, 2}, 2];
v2 = Cone[{{0, 0, 0}, {0, 0, 2*2}}, 2*2*Tan[Pi/3]];
v = RegionIntersection[v1, v2];
Volume[v]

(10Pi)

However, I can't get the result when the radius of the ball and the half apex angle of the cone are variable with the following code:

Clear["Global`*"];
v1 = ImplicitRegion[x^2 + y^2 + (z - a)^2 <= a^2, {x, y, z}]
v2 = ImplicitRegion[z >= Cot[\[Alpha]]*(\!\(TraditionalForm\`
\*SqrtBox[\(
\*SuperscriptBox[\(x\), \(2\)] + 
\*SuperscriptBox[\(y\), \(2\)]\)]\)), {x, y, z}]
v = RegionIntersection[v1, v2];
Assuming[a > 0 && Pi/2 > \!\(TraditionalForm\`\[Alpha]\) > 0, 
 Volume[v]]

($Aborted)

How to get the result?

Answer 1

I doubt Mathematica will find the closed form for that volume with the region given as is, despite it's actually quite simple to derive. I did what @kcr suggested, for the special angles:

LaunchKernels[];
vol = {};
SetSharedVariable[vol];
ParallelDo[
    a = i;
    ang = j;
    v1 = ImplicitRegion[x^2 + y^2 + (z - a)^2 <= a^2, {x, y, z}];
    v2 = ImplicitRegion[z >= Cot[ang]*Sqrt[x^2 + y^2], {x, y, z}];
    v = RegionIntersection[v1, v2];
    AppendTo[vol, {i, ang, Volume[v]}]
, {i, 1, 8}, {j, {Pi/6, Pi/4, Pi/3}}]
CloseKernels[];

vol = Sort[vol]

(* {{1, Pi/6,    (7*Pi)/12}, {1, Pi/4,     Pi}, {1, Pi/3,    (5*Pi)/4},
    {2, Pi/6,   (56*Pi)/12}, {2, Pi/4,   8*Pi}, {2, Pi/3,       10*Pi},
    {3, Pi/6,  (189*Pi)/12}, {3, Pi/4,  27*Pi}, {3, Pi/3,  (135*Pi)/4},
    {4, Pi/6,  (448*Pi)/12}, {4, Pi/4,  64*Pi}, {4, Pi/3,       80*Pi},
    {5, Pi/6,  (875*Pi)/12}, {5, Pi/4, 125*Pi}, {5, Pi/3,  (625*Pi)/4},
    {6, Pi/6,       126*Pi}, {6, Pi/4, 216*Pi}, {6, Pi/3,      270*Pi},
    {7, Pi/6, (2401*Pi)/12}, {7, Pi/4, 343*Pi}, {7, Pi/3, (1715*Pi)/4},
    {8, Pi/6, (3584*Pi)/12}, {8, Pi/4, 512*Pi}, {8, Pi/3,      640*Pi}}  *)

and we know that vol = 4/3*Pi*a^3 for ang >= Pi/4, so what I've got from putting the results for each special angle in FindSequenceFunction is

 7/12*Pi*n^3 (*for Pi/6*)
12/12*Pi*n^3 (*for Pi/4*)
15/12*Pi*n^3 (*for Pi/3*)
16/12*Pi*n^3 (*for Pi/2*)

I put the coefficients again to FindSequenceFunction but no result came out.

The following is the derivation. The intersection simply consists of a cone and a spherical cap who are adjacent at a disk with radius $a\sin(2\alpha)$. Then the height of the cone is $a\sin(2\alpha)\cot(\alpha)$ or, equivalently, $2a\cos^2\alpha$, and that of the cap is $2a\left(1-\cos^2(\alpha)\right)$ or, equivalently, $2a\sin^2(\alpha)$. Put these into the known volume formulas, $$\begin{align*} V_\mathrm{cone}&= \frac{\pi}{3}\left(2a\cos^2\alpha\right)\left(a\sin(2\alpha)\right)^2\\ V_\mathrm{cap}&= \frac{\pi}{6}\left(2a\sin^2(\alpha)\right)\left(3\left(a\sin(2\alpha)\right)^2-(2a\sin^2(\alpha))^2\right)\\ V&=V_\mathrm{cone}+V_\mathrm{cap}=\left(\frac{2}{3}\left(3+\cos(2\alpha)\right)\sin(\alpha)^2\right)\pi a^3 \end{align*}$$

To see this coincides with the computed vol,

A[a_, ang_] := (2/3*(3 + Cos[2*ang])*Sin[ang]^2)*Pi*a^3

Flatten[Table[{i, j, A[i,j]}, {i, 1, 8}, {j, {Pi/6, Pi/4, Pi/3}}], 1] == vol
(* True *)