Volume of a bead with square hole- Program approach?

[Albert Chan]

(11-05-2023 04:28 PM)DM48 Wrote:  d = 12
a = 2
b = 2
...

Volume of sphere solid = 904.779

Volume of sphere with hole removed = 857.0638

Volume of hole removed = 47.715

sphere volume = pi*d^3/6, looks good --> we check removed hole only.

This is a good test of my recent estimate, lo < (average height) < hi

lua> d, a, b = 12, 2, 2
lua> lo = (2*d + sqrt(d*d-a*a-b*b))/3
lua> hi = sqrt(d*d - (a*a+b*b)/3)
lua> a*b*lo, a*b*hi
47.5492050529208      47.55347866700536

Your hole volume is outside this range. sphere (with hole) volume formula is incorrect.
Perhaps you can show steps to derive the formula?

Here is Derive6 calculated hole volume

#1: r := 6
#2: a := 2
#3: b := 2
#4: 8·∫(∫(√(r·r - x·x - y·y), x, 0, a/2), y, 0, b/2)

Shift-Enter (or click [✔≈])      → 47.55262983

Comment:

8 = 2^3, we could scale-up all 3 dimensions, to remove factor 8

#5 d := 2*r
#6: ∫(∫(√(d·d - x·x - y·y), x, 0, a), y, 0, b)      → 47.55262983