When calculating the cross-sectional area of a conductor, S (O'BAC) should refer to the internal bow. ................

Then S (O'BAC)= fan-shaped OBAC- triangle OBC.

Let the angle of BOA be B, then cosb=OO'/OB'/OB, so b=arccosOO'/OB=arccos(t/R).

Sector OBAC=2Rb=Rarccos(t/R), where b takes radian value.

bo'=(ob^2-oo'^2)^( 1/2)=(r^2-t^2)^( 1/2)

So the triangle OBC = 0.5OO' * BC = OO' * BO' = T (R2-T2) (1/2).

So s (o 'bac) = rarccos (t/r)-t (R2-T2) (1/2).