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Der Flächeninhalt von ${\displaystyle {}T}$ ist ${\displaystyle {}\pi /12}$. In Polarkoordinaten wird ${\displaystyle {}T}$ durch den Radius

${\displaystyle {}0\leq r\leq 1\,}$

und den Winkel

${\displaystyle {}0\leq \theta \leq \pi /6\,}$

parametrisiert. Somit ist nach Fakt und Fakt

{\displaystyle {}{\begin{aligned}\int _{T}xd\lambda ^{2}&=\int _{0}^{\pi /6}\int _{0}^{1}r\cos \theta \cdot rdrd\theta \\&={\left(\int _{0}^{1}r^{2}dr\right)}{\left(\int _{0}^{\pi /6}\cos \theta \cdot d\theta \right)}\\&={\frac {1}{3}}\sin \left(\pi /6\right)\\&={\frac {1}{6}}\end{aligned}}}

und

{\displaystyle {}{\begin{aligned}\int _{T}yd\lambda ^{2}&=\int _{0}^{\pi /6}\int _{0}^{1}r\sin \theta \cdot rdrd\theta \\&={\left(\int _{0}^{1}r^{2}dr\right)}{\left(\int _{0}^{\pi /6}\sin \theta \cdot d\theta \right)}\\&={\frac {1}{3}}{\left(1-\cos \left(\pi /6\right)\right)}\\&={\frac {1}{3}}{\left(1-{\frac {\sqrt {3}}{2}}\right)}\\&={\frac {2-{\sqrt {3}}}{6}}.\end{aligned}}}

Somit sind die Koordinaten des Schwerpunktes von ${\displaystyle {}T}$ gleich

${\displaystyle {}\left({\frac {\frac {1}{6}}{\frac {\pi }{12}}},\,{\frac {\frac {2-{\sqrt {3}}}{6}}{\frac {\pi }{12}}}\right)=\left({\frac {2}{\pi }},\,{\frac {4-2{\sqrt {3}}}{\pi }}\right)\,.}$