Integrating over the base, to obtain one-third of the kinetic energy T,
| 1⁄3T = ½ρ ∫ | h / √3 | ¼R2 (3x4 − 1⁄27 h4) dx/h = ρR2 h4 / 135 √3 |
| −h / √3 |
(11)
so that the effective k2 of the liquid filling the triangle is given by
| k2 = T / ½ρR2A = 2h2 / 45 = 2⁄5 (radius of the inscribed circle)2, |
(12)
or two-fifths of the k2 for the solid triangle.
Again, since
dφ/dν = dψ/ds, dφ/ds = −dψ/dν,
(13)