Volumen des gesamten Körpers

Die Skizze zeigt den Axialschnitt $ABCDEFABCDEF$ eines Körpers mit der Rotationsachse $GG$$KK$. Der Punkt $GG$ ist der Schnittpunkt der Geraden $CDCD$ und $FEFE$.

Es gilt: $\overline{AB}=\overline{CF}=5\text{cm}\backslash overline\{AB\}=\backslash overline\{CF\}=5\backslash ;\backslash text\{cm\}$; $\overline{AF}=\overline{BC}=4\text{cm}\backslash overline\{AF\}=\backslash overline\{BC\}=4\backslash ;\backslash text\{cm\}$; $\overline{ED}=\mathrm{2,4}\text{cm}\backslash overline\{ED\}=2\{,\}4\backslash ;\backslash text\{cm\}$; $\mathrm{\sphericalangle }GFK=50\mathrm{°}\backslash sphericalangle\{GFK\}=\; 50°$;

$[AF]\mathrm{\parallel }[GK]\mathrm{\parallel }[BC][AF]\backslash Vert\; [GK]\backslash Vert\; [BC]$.

Das Volumen des Gesamtkörpers (Zylinder) ist: $V={\overline{FK}}^2\cdot \pi \cdot \overline{AF}=\mathrm{78,54}{\text{cm}}^{3}V=\backslash overline\{FK\}^2\backslash cdot\backslash pi\backslash cdot\backslash overline\{AF\}=78\{,\}54~\backslash text\{cm\}^3$

Volumen des Kegels

Berechne die Länge der Strecke $\overline{GK}\backslash overline\{GK\}$:

${\displaystyle \frac{\overline{GK}}{\overline{FK}}}=\tan \mathrm{\sphericalangle }GFK\Rightarrow \overline{GK}=\mathrm{2,5}\text{cm}\cdot \tan {50}^{\circ }\backslash dfrac\{\backslash overline\{GK\}\}\{\backslash overline\{FK\}\}=\backslash tan\backslash sphericalangle\{GFK\}\backslash Rightarrow\backslash overline\{GK\}=2\{,\}5\backslash ;\backslash text\{cm\}\backslash cdot\backslash tan50^\backslash circ$

$\overline{GK}=\mathrm{2,5}\text{cm}\cdot \mathrm{1,1918}\backslash overline\{GK\}=2\{,\}5\backslash ;\backslash text\{cm\}\backslash cdot1\{,\}1918$

$\overline{GK}=\mathrm{2,98}cm\backslash overline\{GK\}=2\{,\}98\backslash \; cm$

Damit ist das Volumen des Kegels: $V={\displaystyle \frac{1}{3}}\cdot {\overline{FK}}^2\cdot \pi \cdot \overline{GK}=\mathrm{19,50}{\text{cm}}^{3}V=\backslash dfrac\{1\}\{3\}\backslash cdot\backslash overline\{FK\}^2\backslash cdot\backslash pi\backslash cdot\backslash overline\{GK\}=19\{,\}50~\backslash text\{cm\}^3$

Volumen der Kegelspitze

Berechne die Länge der Strecke $\overline{GH}\backslash overline\{GH\}$.

${\displaystyle \frac{\overline{GH}}{\overline{GK}}}={\displaystyle \frac{\overline{ED}}{\overline{FC}}}\backslash dfrac\{\backslash overline\{GH\}\}\{\backslash overline\{GK\}\}=\backslash dfrac\{\backslash overline\{ED\}\}\{\backslash overline\{FC\}\}$

$\overline{GH}={\displaystyle \frac{\mathrm{2,98}\text{cm}\cdot \mathrm{2,40}\text{cm}}{5\text{cm}}}\backslash overline\{GH\}=\backslash dfrac\{2\{,\}98\backslash ;\backslash text\{cm\}\backslash cdot2\{,\}40\backslash ;\backslash text\{cm\}\}\{5\backslash ;\backslash text\{cm\}\}$

$\overline{GH}=\mathrm{1,43}\text{cm}\backslash overline\{GH\}=\; 1\{,\}43\backslash ;\backslash text\{cm\}$

Damit ist das Volumen der Kegelspitze: $V={\displaystyle \frac{1}{3}}\cdot {\overline{EH}}^2\cdot \pi \cdot \overline{GH}=\mathrm{2,16}{\text{cm}}^{3}V=\backslash dfrac\{1\}\{3\}\backslash cdot\backslash overline\{EH\}^2\backslash cdot\backslash pi\backslash cdot\backslash overline\{GH\}=2\{,\}16~\backslash text\{cm\}^3$

Gesamtvolumen

Berechne das Volumen des Rotationskörpers.

$V={\overline{FK}}^2\cdot \pi \cdot \overline{AF}-{\displaystyle \frac{1}{3}}\cdot {\overline{FK}}^2\cdot \pi \cdot \overline{GK}+{\displaystyle \frac{1}{3}}\cdot {\overline{EH}}^2\cdot \pi \cdot \overline{GH}V=\backslash overline\{FK\}^2\backslash cdot\backslash pi\backslash cdot\backslash overline\{AF\}-\backslash dfrac\{1\}\{3\}\backslash cdot\backslash overline\{FK\}^2\backslash cdot\backslash pi\backslash cdot\backslash overline\{GK\}+\backslash dfrac\{1\}\{3\}\backslash cdot\backslash overline\{EH\}^2\backslash cdot\backslash pi\backslash cdot\backslash overline\{GH\}$

$V=[(5\cdot \mathrm{0,5}{)}^2\cdot \pi \cdot 4-{\displaystyle \frac{1}{3}}\cdot (5\cdot \mathrm{0,5}{)}^2\cdot \pi \cdot \mathrm{2,98}+{\displaystyle \frac{1}{3}}\cdot (\mathrm{0,5}\cdot \mathrm{2,4}{)}^2\cdot \pi \cdot \mathrm{1,43}]{\text{cm}}^{3}V=[(5\backslash cdot0\{,\}5)^2\backslash cdot\backslash pi\backslash cdot4-\backslash dfrac\{1\}\{3\}\backslash cdot(5\backslash cdot0\{,\}5)^2\backslash cdot\backslash pi\backslash cdot2\{,\}98+\backslash dfrac\{1\}\{3\}\backslash cdot(0\{,\}5\backslash cdot2\{,\}4)^2\backslash cdot\backslash pi\backslash cdot1\{,\}43]\backslash ;\backslash text\{cm\}^3$

$V=\mathrm{61,19}{\text{cm}}^{3}V=61\{,\}19\backslash ;\backslash text\{cm\}^3$