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
$\text{Gegeben:}\\\text{Körperhöhe} \qquad h \qquad [m] \\ \text{Kreiszahl} \qquad \pi \qquad [] \\ \text{Volumen} \qquad V \qquad [m^{3}] \\ \text{Radius 1} \qquad r_{1} \qquad [m] \\ \\ \text{Gesucht:} \\\text{Radius 2} \qquad r_{2} \qquad [m] \\ \\ r_{2} = \sqrt{r_{1} ^{2} - \frac{ V}{\pi \cdot h}}\\ \textbf{Gegeben:} \\ h=4m \qquad \pi=3\frac{16}{113} \qquad V=2m^{3} \qquad r_{1}=6m \qquad \\ \\ \textbf{Rechnung:} \\ r_{2} = \sqrt{r_{1} ^{2} - \frac{ V}{\pi \cdot h}} \\ h=4m\\ \pi=3\frac{16}{113}\\ V=2m^{3}\\ r_{1}=6m\\ r_{2} = \sqrt{6m ^{2} - \frac{ 2m^{3}}{3\frac{16}{113} \cdot 4m}}\\\\r_{2}=5,99m \\\\\\ \small \begin{array}{|l|} \hline h=\\ \hline 4 m \\ \hline 40 dm \\ \hline 400 cm \\ \hline 4\cdot 10^{3} mm \\ \hline 4\cdot 10^{6} \mu m \\ \hline \end{array} \small \begin{array}{|l|} \hline V=\\ \hline 2 m^3 \\ \hline 2\cdot 10^{3} dm^3 \\ \hline 2\cdot 10^{6} cm^3 \\ \hline 2\cdot 10^{9} mm^3 \\ \hline 2\cdot 10^{3} l \\ \hline 20 hl \\ \hline \end{array} \small \begin{array}{|l|} \hline r1=\\ \hline 6 m \\ \hline 60 dm \\ \hline 600 cm \\ \hline 6\cdot 10^{3} mm \\ \hline 6\cdot 10^{6} \mu m \\ \hline \end{array}\\ \small \begin{array}{|l|} \hline r2=\\ \hline 5,99 m \\ \hline 59,9 dm \\ \hline 599 cm \\ \hline 5,99\cdot 10^{3} mm \\ \hline 5,99\cdot 10^{6} \mu m \\ \hline \end{array}$