A={(x,y)| 0≤y≤1 ∧ y-1 ≤ x ≤ √(1-y) }
$$ \int_{0}^{1} \int_{y-1}^{\sqrt{1-y}}1 dxdy $$
$$ =\int_{0}^{1}(\sqrt{1-y} -(y-1)) dy =\frac{7}{6} $$
Zur Kontrolle ohne Doppelintegral:
$$A = 0,5 + \int_{0}^{1} (1-x^2 )dx $$$$A = 0,5 + \frac{2}{3} = \frac{7}{6} $$Passt !
