Doppelintegral/Oberer Halbkreis/x^2y+xy^3/Beispiel

Es sei ${}T$ die obere Einheitskreishälfte und

$f\colon T\longrightarrow \mathbb {R} ,\,(x,y)\longmapsto f(x,y)=x^{2}y+xy^{3}.$

Dann ist nach Fakt

${}{\begin{aligned}\int _{T}fd\lambda ^{2}&=\int _{-1}^{1}{\left(\int _{0}^{\sqrt {1-x^{2}}}x^{2}y+xy^{3}dy\right)}dx\\&=\int _{-1}^{1}{\left({\frac {1}{2}}x^{2}y^{2}+{\frac {1}{4}}xy^{4}\right)}{|}_{0}^{\sqrt {1-x^{2}}}dx\\&=\int _{-1}^{1}{\left({\frac {1}{2}}x^{2}{\sqrt {1-x^{2}}}^{2}+{\frac {1}{4}}x{\sqrt {1-x^{2}}}^{4}\right)}dx\\&=\int _{-1}^{1}{\left({\frac {1}{2}}x^{2}{\left(1-x^{2}\right)}+{\frac {1}{4}}x{\left(1-x^{2}\right)}^{2}\right)}dx\\&=\int _{-1}^{1}{\left({\frac {1}{2}}x^{2}-{\frac {1}{2}}x^{4}+{\frac {1}{4}}x-{\frac {1}{2}}x^{3}+{\frac {1}{4}}x^{5}\right)}dx\\&={\left({\frac {1}{8}}x^{2}+{\frac {1}{6}}x^{3}-{\frac {1}{8}}x^{4}-{\frac {1}{10}}x^{5}+{\frac {1}{24}}x^{6}\right)}{|}_{-1}^{1}\\&={\frac {2}{15}}.\end{aligned}}$