>423
　（補足）V_30の計算

　V_30 = ∫[0,a√(2/3)] (3a^2 -z^2)(z^2)/{(a^2 -z^2)√[(3/4)a^2 -z^2]} dz
　= (27/2)(a^3)∫[(√2 -1)^2, 1] (1+3u^2)(1+(1/3)u^2)(1-u^2)^2 /{(1+14u^2+u^4)(1+u^2)^3} du
　= (a^3)∫[(√2 -1)^2, 1] { 16(1+u^2)/(1+14u^2 +u^4) -(5/2)/(1+u^2) -6/(1+u^2)^2 +6/(1+u^2)^3 } du
　≡ { I - (5/2)J_1 -6J_2 +6J_3 }a^3.

部分分数に分けて、
16(1+u^2)/(1+14u^2+u^4) = 4(2+√3)/(1+(2+√3)^2･u^2) + 4(2-√3)/(1+(2-√3)^2･u^2).
I = 16∫(1+u^2)/(1+14u^2+u^4) du = 4{arctan((2+√3)u) + arctan((2-√3)u)} = 4arctan(4u/(1-u^2)).

J_n ≡∫ {1/(1+u^2)^n} du = {1/(2n-2)}u/(1+u^2)^(n-1) + (2n-3)/(2n-2)}J_(n-1). 　(←部分積分)

∴ I -(5/2)J_1 -6J_2 + 6J_3 = I -(5/2)J_1 -(3/2)J_2 +3u/[2(1+u^2)^2]}
　= I -(13/4)J_1 -3u/[4(1+u^2)] +3u/[2(1+u^2)^2]
　= 4arctan(4u/(1-u^2)) -(13/4)arctan(u) -3u/[4(1+u^2)] +3u/[2(1+u^2)^2].

u: (√2-1)^2 → 1,　1+u^2: 6(√2-1)^2 → 2,　arctan(u): π/4 - (1/2)arccos(1/3) → π/4 より,

V_30 = { I -(5/2)J_1 -6J_2 +6J_3 }a^3
　　= {2π -(29/8)arccos(1/3) -(√2)/12}a^3 = 1.70310628911667･･･(a^3)

V_3 = {2π -(19/4)arccos(1/3) +(√2)/6}a^3 = 0.671830335208･･･a^3.