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# Massenverteilung auf Kreis/x^2y+1/Gesamtmasse/Schwerpunkt/Aufgabe/Lösung

a) Es ist

{\displaystyle {}{\begin{aligned}\int _{P}f(x,y)dxdy&=\int _{-1}^{1}\int _{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}x^{2}y+1dydx\\&=\int _{-1}^{1}{\left({\frac {1}{2}}x^{2}y^{2}+y\right)}{|}_{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}dx\\&=\int _{-1}^{1}2{\sqrt {1-x^{2}}}dx\\&=4\int _{0}^{1}{\sqrt {1-x^{2}}}dx\\&=\pi .\end{aligned}}}

b) Die ${\displaystyle {}x}$-Koordinate des Schwerpunktes muss ${\displaystyle {}0}$ sein, da die Massenverteilung symmetrisch bezüglich der ${\displaystyle {}y}$-Achse (also unter der Spiegelung ${\displaystyle {}x\rightarrow -x}$) ist.

Die ${\displaystyle {}y}$-Koordinate des Schwerpunktes berechnen wir (mit der Substitution ${\displaystyle {}x=\sin u}$) zu

{\displaystyle {}{\begin{aligned}y_{S}&={\frac {1}{\pi }}\int _{P}yf(x,y)dxdy\\&={\frac {1}{\pi }}\int _{-1}^{1}\int _{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}x^{2}y^{2}+ydydx\\&={\frac {1}{\pi }}\int _{-1}^{1}{\left({\frac {1}{3}}x^{2}y^{3}+{\frac {1}{2}}y^{2}\right)}{|}_{-{\sqrt {1-x^{2}}}}^{\sqrt {1-x^{2}}}dx\\&={\frac {1}{\pi }}\int _{-1}^{1}{\frac {2}{3}}x^{2}{\left(1-x^{2}\right)}{\sqrt {1-x^{2}}}dx\\&={\frac {4}{3\pi }}\int _{0}^{1}x^{2}{\left(1-x^{2}\right)}{\sqrt {1-x^{2}}}dx\\&={\frac {4}{3\pi }}\int _{0}^{\pi /2}{\left(\sin ^{2}u\right)}{\left(1-\sin ^{2}u\right)}{\left(\cos u\right)}{\left(\cos u\right)}du\\&={\frac {4}{3\pi }}\int _{0}^{\pi /2}\sin ^{2}u-2\sin ^{4}u+\sin ^{6}udu\\&={\frac {4}{3\pi }}{\left({\frac {1}{4}}\pi -{\frac {3}{8}}\pi +{\frac {5}{32}}\pi \right)}\\&={\frac {1}{3}}-{\frac {1}{2}}+{\frac {5}{24}}\\&={\frac {1}{24}}.\end{aligned}}}