We need moments 0, 1 and 2 of our blob. For this we need the moments of a sliver with its point at the origin. For a triangular surface element, let D be ½ the determinant of the coördinates of the three vertices. For a sliver whose big end is the centroid at (X⁰, X¹, X²) of those vertices we have:
M = D∫[0≤λ≤1]λ²dλ = D/3.
M^j = D∫[0≤λ≤1](λX^j)λ²dλ = D X^j/4.
M^jk = D∫[0≤λ≤1](λX^j)(λX^k)λ²dλ = D X^jX^k/5.
N.B. j and k are coördinate identifiers, not exponents.

The block of code that includes the comment “turn” comes from this logic.

From this code which randomly samples one octant of a the unit sphere, I conclude that for the whole sphere
M ≃ 4.189 (4/3 π ≃ 4.1888)
and M⁰⁰ ≃ 0.8379 (4/15 π ≃ 0.8378)

For what its worth we also learn that for the hemisphere where x>0,
M⁰ ≃ 0.785; M = 2.0944 and centroid is a (M⁰/M, 0, 0) = (3.746, 0, 0)