= ½ρ ∫ φ ds = ½ρ ∫ ψ ds.

(7)

For example, in the equilateral triangle of (8) § 28, referred to coordinate axes made by the base and height,

ψ′ = −2Rαβγ/h = −½ Ry [ (h − y)2 − 3x2 ] /h

(8)

ψ = ψ′ − ½R [ ( 1⁄3 h − y)2 + x2 ] = −½R [ ½h3 + 1⁄3 h2y + h) (x2 − y2) − 3x2y + y3 ] /h

(9)

and over the base y = 0,

dx/dν = −dx/dy = + ½R ( 1⁄3 h2 − 3x2) / h, ψ = −½R ( 1⁄9 h2 + x2).

(10)