Beispiel Nr: 01
$\text{Gegeben:}\\\text{Körperhöhe} \qquad h \qquad [m] \\ \text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Radius 2} \qquad r_{2} \qquad [m] \\ \text{Volumen} \qquad V \qquad [m^{3}] \\ \\ \text{Gesucht:} \\\text{Radius 1} \qquad r_{1} \qquad [m] \\ \\ r_{1} = \sqrt{\frac{ V}{\pi \cdot h}+r_{2} ^{2} }\\ \textbf{Gegeben:} \\ h=4m \qquad \pi=3,14 \qquad r_{2}=5m \qquad V=6m^{3} \qquad \\ \\ \textbf{Rechnung:} \\ r_{1} = \sqrt{\frac{ V}{\pi \cdot h}+r_{2} ^{2} } \\ h=4m\\ \pi=3,14\\ r_{2}=5m\\ V=6m^{3}\\ r_{1} = \sqrt{\frac{ 6m^{3}}{3,14 \cdot 4m}+5m ^{2} }\\\\r_{1}=5,05m \\\\\\ \tiny \begin{array}{|l|} \hline h=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline 0,004 km \\ \hline \end{array} \tiny \begin{array}{|l|} \hline r2=\\ \hline 5 m \\ \hline 50 dm \\ \hline 500 cm \\ \hline 5\cdot 10^{3} mm \\ \hline 5\cdot 10^{6} \mu m \\ \hline 0,005 km \\ \hline \end{array} \tiny \begin{array}{|l|} \hline V=\\ \hline 6 m^3 \\ \hline 6\cdot 10^{3} dm^3 \\ \hline 6\cdot 10^{6} cm^3 \\ \hline 6\cdot 10^{9} mm^3 \\ \hline 6\cdot 10^{3} l \\ \hline 60 hl \\ \hline 6\cdot 10^{6} ml \\ \hline \end{array}\\ \tiny \begin{array}{|l|} \hline r1=\\ \hline 5,05 m \\ \hline 50,5 dm \\ \hline 505 cm \\ \hline 5,05\cdot 10^{3} mm \\ \hline 5,05\cdot 10^{6} \mu m \\ \hline 0,00505 km \\ \hline \end{array}$