Difference between revisions of "MR 07 Loesung"
| Line 30: | Line 30: | ||
<math>2 \pi \left ( {h \over 2} \cdot {{h^2} \over 4} - \left | {{x^3} \over 3} \right |_0^{h \over 2} \right )</math> | <math>2 \pi \left ( {h \over 2} \cdot {{h^2} \over 4} - \left | {{x^3} \over 3} \right |_0^{h \over 2} \right )</math> | ||
| − | <math>2 \pi \left ( {{h^3} \over 8} - { {{ \left ( {h \over 2} \right ) }^3} \over 3} \right ) | + | <math>2 \pi \left ( {{h^3} \over 8} - { {{ \left ( {h \over 2} \right ) }^3} \over 3} \right ) = 2 \pi \left ( \left ( {h \over 2} \right )^3 - {1 \over 3} \cdot \left ( {h \over 2} \right )^3 \right ) = 2 \pi \left ( {2 \over 3} \cdot \left ( {h \over 2} \right )^3 \right )</math> |
| − | |||
| − | |||
<math>{{4 \pi} \over 3} \cdot \left ( {h \over 2} \right )^3</math> | <math>{{4 \pi} \over 3} \cdot \left ( {h \over 2} \right )^3</math> | ||
Und das ist das Volumen einer Kugel mit dem Durchmesser <math>h</math>! | Und das ist das Volumen einer Kugel mit dem Durchmesser <math>h</math>! | ||
Revision as of 15:35, 18 September 2018
Die Perle des Abakus
Die Skizze hat ein relativ großes Loch, weil man so die Verhältnisse besser erkennen kann:
Durch das (gebohrte) Loch hat die Kugel die Deckkalotten verloren. Die Kugeln auf der Stange der Kugerlrechnenmaschine sitzen näher beieinander als der Durchmesser Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle D} der Kugel vorgibt. Dieses kürzere Maß nenne ich Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} und es hängt natürlich von Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D} und Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} ab:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h = \sqrt{D^2 - d^2}}
Um das gesuchte Volumen der Kugel mit Durchmesser Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D} mit dem Loch mit Durchmesser Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle d} zu finden führt (in diesem Fall) der einfachste und schnellste Weg über ein Integral der Querschnitte. Es wird von Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0} bis Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h \over 2} integriert und das Ergebnis mal zwei genommen um das Gesamtvolumen zu erhalten, das auch aus der anderen Hälfte besteht.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \int\limits_{0}^{h \over 2}\pi \left (\sqrt{{{D^2} \over 4} - x^2} \right )^2 - \pi {{d^2} \over 4}\, dx}
Das mit der Wurzel ist der Radius der Kreisscheibe der Kugel an der Stelle x. Natürlich kann man die Wurzel gegen das Quadrat "kürzen" - genauso, wie Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} herausheben und vor das Integral stellen.
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi \int\limits_{0}^{h \over 2} {{D^2} \over 4} - x^2 - {{d^2} \over 4} \, dx}
Ein wenig umrangieren:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi \int\limits_{0}^{h \over 2} {{D^2 - d^2} \over 4} - x^2 \, dx}
Und man erkennt das Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle D^2 - d^2} als Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h^2} .
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi \int\limits_{0}^{h \over 2} {{h^2} \over 4} - x^2 \, dx}
Man kann das ganze auch in zwei Integrale zerlegen, die sich einzeln leichter auflösen lassen:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi \left ( \int\limits_{0}^{h \over 2} {{h^2} \over 4} - \int\limits_{0}^{h \over 2} x^2 \, dx \right )}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi \left ( {h \over 2} \cdot {{h^2} \over 4} - \left | {{x^3} \over 3} \right |_0^{h \over 2} \right )}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2 \pi \left ( {{h^3} \over 8} - { {{ \left ( {h \over 2} \right ) }^3} \over 3} \right ) = 2 \pi \left ( \left ( {h \over 2} \right )^3 - {1 \over 3} \cdot \left ( {h \over 2} \right )^3 \right ) = 2 \pi \left ( {2 \over 3} \cdot \left ( {h \over 2} \right )^3 \right )}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {{4 \pi} \over 3} \cdot \left ( {h \over 2} \right )^3}
Und das ist das Volumen einer Kugel mit dem Durchmesser Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h} !