Volume of a bead with square hole- Program approach?

[Albert Chan]

(06-11-2020 06:53 PM)Albert Chan Wrote:  Solve with A=0.1, B=0.1, we have hole = 0.00996658845699, about 2% of sphere

For small square hole, we can estimate with simple formula
For unit diameter (d=1), with square hole, b=a

XCas> k := sqrt(1-2*b*b)
XCas> hv2 := b*b*k/3 - atan(b*b/k)/3 + b*(1-b*b/3)*atan(b/k)
XCas> hv2(b = 0.1)             // just to confirm square hole formula
→ 0.00996658845699

XCas> taylor(hv2,b,10)
→ b^2 - b^4/3 + -7*b^6/90 + -3*b^8/70 + -83*b^10/2520 + b^12*order_size(b)

Since taylor series all have even powers, let dimensionless x = b^2

XCas> pade(x - x^2/3 - 7*x^3/90, x, 3,3)
→ (17*x^2-30*x)/(7*x-30)

XCas> r(x) := x*(17*x-30)/(7*x-30)
XCas> r(0.1^2)                     // rough estimate for small 0.1×0.1 hole
→ 0.00996658870698