Aloha :)
$$I=2\pi\int\limits_0^3\frac{1}{3}x^3\sqrt{1+\frac{x^4}{81}}\,dx=\frac{2\pi}{3}\int\limits_0^3x^3\sqrt{1+\left(\frac{x}{3}\right)^4}\,dx$$Wir substituieren wie folgt:$$u:=\left(\frac{x}{3}\right)^4=\frac{x^4}{81}\;\;;\;\;u(0)=0\;\;;\;\;u(3)=1\;\;;\;\;\frac{du}{dx}=\frac{4x^3}{81}\;\Rightarrow\;dx=\frac{81\,du}{4x^3}$$$$I=\frac{2\pi}{3}\int\limits_0^1x^3\sqrt{1+u}\,\frac{81\,du}{4x^3}=\frac{27}{2}\pi\int\limits_0^1\sqrt{1+u}\,du$$Wir substituieren erneut:$$y=1+u\quad;\quad y(0)=1\quad;\quad y(1)=2\quad;\quad dy=du$$$$I=\frac{27}{2}\pi\int\limits_1^2\sqrt y\, dy=\frac{27}{2}\pi\int\limits_1^2y^{1/2}\, dy=\frac{27}{2}\pi\left[\frac{2}{3}y^{3/2}\right]_1^2=9\pi\left(2\sqrt2-1\right)\approx51,698$$
