Geometrie-Stereometrie-Hohlzylinder

$V = (r_{1} ^{2} - r_{2} ^{2} )\cdot \pi \cdot h$
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$r_{1} = \sqrt{\frac{ V}{\pi \cdot h}+r_{2} ^{2} }$
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$r_{2} = \sqrt{r_{1} ^{2} - \frac{ V}{\pi \cdot h}}$
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$h = \frac{ V}{(r_{1} ^{2} - r_{2} ^{2} )\cdot \pi }$
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Beispiel Nr: 01
$\begin{array}{l} \text{Gegeben:}\\\text{Volumen} \qquad V \qquad [m^{3}] \\ \text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Radius 2} \qquad r_{2} \qquad [m] \\ \text{Radius 1} \qquad r_{1} \qquad [m] \\ \\ \text{Gesucht:} \\\text{Körperhöhe} \qquad h \qquad [m] \\ \\ h = \frac{ V}{(r_{1} ^{2} - r_{2} ^{2} )\cdot \pi }\\ \textbf{Gegeben:} \\ V=4m^{3} \qquad \pi=3\frac{16}{113} \qquad r_{2}=2m \qquad r_{1}=6m \qquad \\ \\ \textbf{Rechnung:} \\ h = \frac{ V}{(r_{1} ^{2} - r_{2} ^{2} )\cdot \pi } \\ V=4m^{3}\\ \pi=3\frac{16}{113}\\ r_{2}=2m\\ r_{1}=6m\\ h = \frac{ 4m^{3}}{(6m ^{2} - 2m ^{2} )\cdot 3\frac{16}{113} }\\\\h=0,0398m \\\\\\ \small \begin{array}{|l|} \hline V=\\ \hline 4 m^3 \\ \hline 4\cdot 10^{3} dm^3 \\ \hline 4\cdot 10^{6} cm^3 \\ \hline 4\cdot 10^{9} mm^3 \\ \hline 4\cdot 10^{3} l \\ \hline 40 hl \\ \hline \end{array} \small \begin{array}{|l|} \hline r2=\\ \hline 2 m \\ \hline 20 dm \\ \hline 200 cm \\ \hline 2\cdot 10^{3} mm \\ \hline 2\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline r1=\\ \hline 6 m \\ \hline 60 dm \\ \hline 600 cm \\ \hline 6\cdot 10^{3} mm \\ \hline 6\cdot 10^{6} \mu m \\ \hline \end{array}\\ \small \begin{array}{|l|} \hline h=\\ \hline 0,0398 m \\ \hline 0,398 dm \\ \hline 3,98 cm \\ \hline 39,8 mm \\ \hline 3,98\cdot 10^{4} \mu m \\ \hline \end{array} \end{array}$