Verfahren "wie?": (1,0) wählen, dann 2. Punkt, dann 3. Punkt (wie döschwo).

A(Dreieck)
= 1/2 * I (eix - e0) x ((eiy - e0) I
=1/2 * I [ ⎝⎛cosxsinx0⎠⎞ - ⎝⎛100⎠⎞ ] x [⎝⎛cosysiny0⎠⎞ - ⎝⎛100⎠⎞ ] I
= 1/2 * I ⎝⎛cosx−1sinx0⎠⎞ x ⎝⎛cosy−1siny0⎠⎞ I
= 1/2 * I ⎝⎛00(cosx−1)siny−sinx(cosy−1)⎠⎞ I
= 1/2 * I (cos x - 1)sin y - sin x (cos y - 1) I
Mittelwert(A(Dreieck))
= 21 * 2π∗2π1 0∫2π 0∫2π I (cos x - 1)sin y - sin x (cos y - 1) I dx dy *) Nebenrechnung
= - 21 * 2π∗2π1 0∫2π y∫2π(cos x - 1)sin y - sin x (cos y - 1) dx dy
+ 21 * 2π∗2π1 0∫2π 0∫y(cos x - 1)sin y - sin x (cos y - 1) dx dy
=8π21[ 0∫2π0∫y(cos x -1)sin y - sin x (cos y -1) dxdy - 0∫2πy∫2π(cos x -1)sin y - sin x (cos y -1) dxdy]
= 8π21 [ 0∫2π 0∫y sin(y - x) + sin(x) - sin(y) dx dy - 0∫2π y∫2π sin(y - x) + sin(x) - sin(y) dx dy ]
= 8π21 [0∫2π[cos(y - x) - cos(x) - x sin(y) ] (0..y) dy - 0∫2π[cos(y - x) - cos(x) - x sin(y) ] (y..2π) dy]
= 8π21 [ 0∫2π 1-cos(y) - y sin(y) -cos(y) +1 dy - 0∫2π cos(y) -1 -0 -1 + cos (y) +y sin(y) dy ]
= 8π21 [ 0∫2π 2-2cos(y) - y sin(y) dy - 0∫2π 2cos(y) -2 +y sin(y) dy ]
= 4π21 0∫2π 2-2cos(y) - y sin(y) dy
= 4π21 [ 2 y - 3 sin(y) + y cos(y) ] (0..2π)
= 4π21 [ 4π + 2π ]
=1,5 / π
= 0,4774648 ....
Nebenrechnung:
(cos x - 1)sin y - sin x (cos y - 1)
= -sin(x - y) + sin(x) - sin(y)
= sin(y - x) + sin(x) - sin(y)
= 4 sin(x/2) sin(y/2) sin(y/2 - x/2) jeweils 4π-periodisch
≥0 für y≥x
{
<0 sonst