HP42s first major program (Double Integral) Best way to approach?

[Albert Chan]

(06-06-2020 08:33 PM)Albert Chan Wrote:  

(06-02-2020 08:31 AM)Werner Wrote:  Question: wouldn't the formula need to be

bore hole volume = hv(b,c) - hv(max(a,c),min(a,c))

so that it would work for a<c as well?

Amazingly, the answer is NO !
...
The real part returns the same volume !

We can prove this with Legendre's relation, 19.7.3
Note: the prove use elliptical parameter m instead of modulus, k, m = k²

Assume d < D = 1, so that m = (d/D)² = d² < 1
(we can scale it back later, by factor D³)

3*HV(d,1) = m*(E+K) + (E-K) = (m+1)*E + (m-1)*K

With order flipped, using notation K(1/m) = K(m) , K(1-m) = K'(m)

3*HV(1,d)
= d³ * 3*HV(1/d,1)
= d³ * ((1/m+1)*E + (1/m-1)*K)
= (m+1) * (d*E) + (m-m²) * (K/d)
= (m+1) * (E - (1-m)*K ± 1j * (E' - m*K')) + (m-m²) * (K ∓ 1j * K')

Re(3*HV(1,d)) = (m+1)*E + (m²-1)*K + (m-m²)*K = 3*HV(d,1)

Adding the special case, HV(1,1) = 2/3, and removed assumptions d < D = 1:

hole volume = Re(HV(d,D)) = Re(HV(D,d))