10_Rotationskoerper_EbeKue.docx
Lösungen zu Rotationskörpern
Aufgabe 1:
= + + +
=1 2 ∙ 4
3 ∙ 15 !∙ "# + 15 ∙ " ∙ 5 + 5 ∙ " ∙ 30 +1
3 ∙ 5 ∙ " ∙ 10
=2
3 ∙ 3375 !∙ " + 1125 !∙ " + 750 !∙ " +1
3 ∙ 250 !∙ "
≈ 2250 !∙ " + 1125 !∙ " + 750 !∙ " + 8,3 !∙ "
= )2250 + 1125 + 750 + 8,3* ∙ " !
= 4133,3 ∙ " !≈ 12978,56 !≈ 12,98 -!
Aufgabe 2:
ö/ = + − Zylinder mit Grundkreisradius a sowie Höhe 2a
=1
3 ⋅ 2⋅ " ⋅ 2 + 2⋅ " ⋅ 22 −1 2 ⋅ 4
3 ⋅ 2!⋅ "#
=1
3 ⋅ 2!⋅ " + 22!⋅ " −2
3 ⋅ 2!⋅ "
= 1
3 + 2 −2
3# ⋅ 2!⋅ "
= 12
3 ⋅ 2!⋅ "
3ö/ = 4 + 4 + 3 Zylinder mit Grundkreisradius 2 und Höhe 22
= ⋅ 2 ⋅ " + )22"* ⋅ 22 +⋅ )42⋅ "* Kegel mit Mantellinie m: = 2+ 2⇔ = √2 ⋅ 2
= √2 ⋅ 2 ⋅ 27
8 ⋅ " + 42" + 22⋅ "
= 9√2 + 4 + 2:2⋅ " = ;6 + √2<2⋅ "
10_Rotationskoerper_EbeKue.docx Aufgabe 3:
a) Skizze
b) ö/ = 2 ⋅ = 2 ⋅ =!⋅ 8 - ⋅ " ⋅ 8 -> ≈ 1072,33 -!
c) Der Rotationskörper besteht aus zwei aufeinander stehenden Kegeln und enthält eine Hohlkugel.
ö/ = ö/ − = 2 ⋅ 1
3 ⋅ 8 - ⋅ " ⋅ 8 -# −4
3 ⋅ 5- !⋅ " ≈ 548,73 -!
d) Anteil des Volumens:
ö/
ö/ = 548,73 -! 1072,33 -!≈ 0,5
Das Volumen des Körpers verringert sich in etwa um die Hälfte (50 %).
