Kreissektor/45 Grad/Schwerpunkt/Polarkoordinaten/Aufgabe/Lösung

Der Flächeninhalt von ${}T$ ist ${}\pi /8$ . In Polarkoordinaten wird ${}T$ durch den Radius

{}0\leq r\leq 1\,

und den Winkel

{}0\leq \theta \leq \pi /4\,

parametrisiert. Somit ist nach Fakt und Fakt

{}{\begin{aligned}\int _{T}xd\lambda ^{2}&=\int _{0}^{\pi /4}\int _{0}^{1}r\cos \theta \cdot rdrd\theta \\&={\left(\int _{0}^{1}r^{2}dr\right)}{\left(\int _{0}^{\pi /4}\cos \theta \cdot d\theta \right)}\\&={\frac {1}{3}}\sin \left(\pi /4\right)\\&={\frac {1}{3}}\cdot {\frac {1}{\sqrt {2}}}\\&={\frac {1}{3{\sqrt {2}}}}\end{aligned}}

und

{}{\begin{aligned}\int _{T}yd\lambda ^{2}&=\int _{0}^{\pi /4}\int _{0}^{1}r\sin \theta \cdot rdrd\theta \\&={\left(\int _{0}^{1}r^{2}dr\right)}{\left(\int _{0}^{\pi /4}\sin \theta \cdot d\theta \right)}\\&={\frac {1}{3}}{\left(1-\cos \left(\pi /4\right)\right)}\\&={\frac {1}{3}}{\left(1-{\frac {1}{\sqrt {2}}}\right)}\\&={\frac {{\sqrt {2}}-1}{3{\sqrt {2}}}}.\end{aligned}}

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

{}\left({\frac {\frac {1}{3{\sqrt {2}}}}{\frac {\pi }{8}}},\,{\frac {\frac {{\sqrt {2}}-1}{3{\sqrt {2}}}}{\frac {\pi }{8}}}\right)=\left({\frac {8}{3{\sqrt {2}}\pi }},\,{\frac {8({\sqrt {2}}-1)}{3{\sqrt {2}}\pi }}\right)\,.