Rotationskörper/0 bis 1/t+sqrt(t)+1/Aufgabe/Lösung

Das Volumen des Rotationskörpers ${}K$ ist gemäß der Formel gleich

{}{\begin{aligned}\lambda ^{3}(K)&=\pi \int _{0}^{1}(t+{\sqrt {t}}+1)^{2}\,dt\\&=\pi \int _{0}^{1}(t^{2}+t+1+2t^{3/2}+2t+2t^{1/2})\,dt\\&=\pi \int _{0}^{1}(t^{2}+2t^{3/2}+3t+2t^{1/2}+1)\,dt\\&=\pi {\left({\frac {1}{3}}t^{3}+{\frac {4}{5}}t^{5/2}+{\frac {3}{2}}t^{2}+{\frac {4}{3}}t^{3/2}+t\right)}|_{0}^{1}\\&=\pi {\left({\frac {1}{3}}+{\frac {4}{5}}+{\frac {3}{2}}+{\frac {4}{3}}+1\right)}\\&=\pi {\frac {10+24+45+40+30}{30}}\\&=\pi {\frac {149}{30}}.\end{aligned}}