Aloha :)
Entlang des Weges$$\vec r=\begin{pmatrix}2t\cos t-2\sin t\\2t\sin t+2\cos t\\4t^2\end{pmatrix}\quad;\quad t\in[0;2\pi]$$wir der Integrand zu:$$x^2+y^2+z^2$$$$\quad=4t^2\cos^2t-8t\sin t\cos t+4\sin^2 t+4t^2\sin^2t+8t\sin t\cos t+4\cos^2t+16t^4$$$$\quad=4t^2+4+16t^4=4(4t^4+t^2+1)$$und das Linienelement zu:$$ds=\left\|\frac{d\vec r}{dt}\right\|dt=\left\|\begin{pmatrix}2\cos t-2t\sin t-2\cos t\\2\sin t+2t\cos t-2\sin t\\8t\end{pmatrix}\right\|dt=\left\|\begin{pmatrix}-2t\sin t\\2t\cos t\\8t\end{pmatrix}\right\|dt$$$$\phantom{ds}=2t\left\|\begin{pmatrix}-\sin t\\\cos t\\4\end{pmatrix}\right\|dt=2t\sqrt{17}\,dt$$
Damit haben wir das gesuchte Kurvenintegral$$I=\int\limits_0^{2\pi}4(4t^4+t^2+1)\cdot2t\sqrt{17}\,dt=8\sqrt{17}\int\limits_0^{2\pi}\left(4t^5+t^3+t\right)dt$$$$\phantom{I}=8\sqrt{17}\left[\frac{2t^6}{3}+\frac{t^4}{4}+\frac{t^2}{2}\right]_0^{2\pi}=\frac{8}{12}\sqrt{17}\left[8t^6+3t^4+6t^2\right]_0^{2\pi}$$$$\phantom{I}=\frac{16\sqrt{17}}{3}\,\pi^2\left(3+6\pi^2+64\pi^4\right)$$
