A 0 = α − 1 π ∫ 0 π d Z ¯ d x ( θ ) d θ {\displaystyle A_{0}=\alpha -{\frac {1}{\pi }}\int _{0}^{\pi }{\frac {d{\overline {Z}}}{dx}}(\theta )d\theta }
A n = 2 π ∫ 0 π d Z ¯ d x ( θ ) cos ( n θ ) d θ {\displaystyle A_{n}={\frac {2}{\pi }}\int _{0}^{\pi }{\frac {d{\overline {Z}}}{dx}}(\theta )\cos(n\theta )d\theta }
L ( x ) = ∫ 0 x w ( x ∗ ) d x ∗ = ∫ 0 x c 1 − 4 x ∗ 2 b 2 d x ∗ = 1 / 2 c x 1 − 4 x 2 b 2 + 1 / 4 c arctan ( 2 b − 2 x 1 1 − 4 x 2 b 2 ) 1 b − 2 {\displaystyle L(x)=\int _{0}^{x}w(x^{*})dx^{*}=\int _{0}^{x}c{\sqrt {1-{\frac {4x^{*2}}{b^{2}}}}}dx^{*}=1/2\,cx{\sqrt {1-4\,{\frac {{x}^{2}}{{b}^{2}}}}}+1/4\,c\arctan \left(2\,{\sqrt {{b}^{-2}}}x{\frac {1}{\sqrt {1-4\,{\frac {{x}^{2}}{{b}^{2}}}}}}\right){\frac {1}{\sqrt {{b}^{-2}}}}}
where:
c e n t r o i d = ∫ 0 x x ∗ c 1 − 4 x ∗ 2 b 2 d x ∗ ∫ 0 x c 1 − 4 x ∗ 2 b 2 d x ∗ = 1 / 12 ( 2 x − b ) ( 2 x + b ) c − − b 2 + 4 x 2 b 2 ( 1 / 2 c x 1 − 4 x 2 b 2 + 1 / 4 c arctan ( 2 b − 2 x 1 1 − 4 x 2 b 2 ) 1 b − 2 ) − 1 {\displaystyle \mathrm {centroid} ={\frac {\int _{0}^{x}x^{*}c{\sqrt {1-{\frac {4x^{*2}}{b^{2}}}}}dx^{*}}{\int _{0}^{x}c{\sqrt {1-{\frac {4x^{*2}}{b^{2}}}}}dx^{*}}}=1/12\,\left(2\,x-b\right)\left(2\,x+b\right)c{\sqrt {-{\frac {-{b}^{2}+4\,{x}^{2}}{{b}^{2}}}}}\left(1/2\,cx{\sqrt {1-4\,{\frac {{x}^{2}}{{b}^{2}}}}}+1/4\,c\arctan \left(2\,{\sqrt {{b}^{-2}}}x{\frac {1}{\sqrt {1-4\,{\frac {{x}^{2}}{{b}^{2}}}}}}\right){\frac {1}{\sqrt {{b}^{-2}}}}\right)^{-1}}
is the total force at point x
is the span
is the integration variable (kinda like how sometimes you integrate in 
from 0 to t)
