Geometrie-Stereometrie-Hohlzylinder


  • $V = (r_{1} ^{2} - r_{2} ^{2} )\cdot \pi \cdot h$
    1 2
    $r_{1} = \sqrt{\frac{ V}{\pi \cdot h}+r_{2} ^{2} }$
    1
    $r_{2} = \sqrt{r_{1} ^{2} - \frac{ V}{\pi \cdot h}}$
    1
    $h = \frac{ V}{(r_{1} ^{2} - r_{2} ^{2} )\cdot \pi }$
    1

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\frac{16}{113} \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\frac{16}{113}\\ r_{2}=5m\\ V=6m^{3}\\ r_{1} = \sqrt{\frac{ 6m^{3}}{3\frac{16}{113} \cdot 4m}+5m ^{2} }\\\\r_{1}=5,05m \\\\\\ \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 r2=\\ \hline 5 m \\ \hline 50 dm \\ \hline 500 cm \\ \hline 5\cdot 10^{3} mm \\ \hline 5\cdot 10^{6} \mu m \\ \hline \end{array} \small \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 \end{array}\\ \small \begin{array}{|l|} \hline r1=\\ \hline 5,05 m \\ \hline 50,5 dm \\ \hline 504\frac{91}{121} cm \\ \hline 5047\frac{63}{121} mm \\ \hline 5,05\cdot 10^{6} \mu m \\ \hline \end{array}$