I think you are right, I just was too quickie in reading the question.
Your answer is correct.
The body of the solid I envision, it has just the shape of a that kind of multipurpose plug that can in the puzzle books and it can be used to close three different kind of holes circular, isosceles triangle or a rectangular hole, located inside a crystal box the three main views are a circle, a rectangle and a triangle.
It has the very same shape of a cylinder if we remove two cylindrical wedges. So its volume will be the cylinder volume Vc = pi*Hm*R^2 , less the two wedges volume. Vw= 2*Hm*R^2 times two .
Which amount obviously to Vn= Hm*R^2(pi-4/3)
Making Hm=20, and R=5.
You get the nine hundred and four thousand... you posted. The only inacuracy in your formula is pi/4 should be pi/2.
For me using calculus, is the easier way to go, probably Archimedes got it without it, and maybe we can get it too there,.. it will take a lot of ingenuity.
Using calculus the answer is almost immediate, we can calculate either the half of our solid or even easier the cylindrical wedges.
To obtain the volume either the solid or the wedges, in both cases the trick is slicing the solid vertically and parallel to the given diameter we obtain rectangular sections for the cylindrical wedges, below it is the calculus for each wedge
Doing the slicing as indicated a general section at distance r of the diameter will be
S(r) = (Hm/R))*r*(R^2-r^2)^.5
We must integrate S(r) between 0 and R.
I know you can do it close eyed, either directly, using trig or even easier making the change w=(r/R)^2 , upper limit 1; lower limit 0...
The answer for each wedge is Vw=2/3 Hm*R^2
But I am still thinking in how it might be obtained without the calculus help.