Für diese Aufgabe benötigst Du folgendes Grundwissen: Zylinder
Teillösung 1: Volumen Maße großer Zylinder: d 1 = 6 cm d_1=6\;\text{cm} d 1 = 6 cm h 1 = 3 , 5 cm h_1=3{,}5\;\text{cm} h 1 = 3 , 5 cm
Der Radius ist halb so lang wie der Durchmesser ⇒ r 1 = d 1 2 = 3 cm \Rightarrow r_1=\dfrac{d_1}{2}=3\;\text{cm} ⇒ r 1 = 2 d 1 = 3 cm
Maße kleiner Zylinder: d 2 = 2 cm d_2=2\;\text{cm} d 2 = 2 cm h 2 = 4 cm h_2=4\;\text{cm} h 2 = 4 cm
Der Radius ist halb so lang wie der Durchmesser ⇒ r 2 = d 2 2 = 1 cm \Rightarrow r_2=\dfrac{d_2}{2}=1\;\text{cm} ⇒ r 2 = 2 d 2 = 1 cm
Das Volumen eines Zylinders berechnest du mit der Formel:
V Z y l i n d e r = r 2 ⋅ π ⋅ h k V_{Zylinder} = r^2\cdot\pi\cdot h_k V Z y l in d er = r 2 ⋅ π ⋅ h k
V Z \displaystyle V_Z V Z = = = V g r o ß e r Z y l i n d e r + V k l e i n e r Z y l i n d e r \displaystyle V_{großerZylinder}+V_{kleinerZylinder} V g ro ß er Z y l in d er + V k l e in er Z y l in d er = = = r 1 2 ⋅ π ⋅ h 1 + r 2 2 ⋅ π ⋅ h 2 \displaystyle r_1^2\cdot\pi\cdot h_1+ r_2^2\cdot \pi\cdot h_2 r 1 2 ⋅ π ⋅ h 1 + r 2 2 ⋅ π ⋅ h 2 = = = ( 3 c m ) 2 ⋅ π ⋅ 3 , 5 c m + ( 1 c m ) 2 ⋅ π ⋅ 4 c m \displaystyle \mathrm{(3\;cm)}^2\cdot\pi\cdot \mathrm{3{,}5\;cm}+ \mathrm{(1\;cm)}^2\cdot \pi\cdot \mathrm{4\;cm} ( 3 cm ) 2 ⋅ π ⋅ 3 , 5 cm + ( 1 cm ) 2 ⋅ π ⋅ 4 cm ↓
π ≈ 3 , 14 \pi \approx 3{,}14 π ≈ 3 , 14
≈ ≈ ≈ ( 3 c m ) 2 ⋅ 3 , 14 ⋅ 3 , 5 c m + ( 1 c m ) 2 ⋅ 3 , 14 ⋅ 4 c m \displaystyle \mathrm{(3\;cm)}^2\cdot3{,}14\cdot \mathrm{3{,}5\;cm}+ \mathrm{(1\;cm)}^2\cdot 3{,}14\cdot \mathrm{4\;cm} ( 3 cm ) 2 ⋅ 3 , 14 ⋅ 3 , 5 cm + ( 1 cm ) 2 ⋅ 3 , 14 ⋅ 4 cm = = = 98 , 91 c m 3 + 12 , 56 c m 3 \displaystyle 98{,}91\mathrm{\;cm}^3+12{,}56 \mathrm{\;cm}^3 98 , 91 cm 3 + 12 , 56 cm 3 = = = 111 , 47 c m 3 \displaystyle 111{,}47\mathrm{\;cm}^3 111 , 47 cm 3
Antwort: Das Werkstück hat ein Volumen von etwa 111 , 47 c m 3 111{,}47\mathrm{\;cm}^3 111 , 47 cm 3
Teillösung 2: Oberflächeninhalt Die Oberfläche dieses Werkstücks besteht aus mehreren Teilflächen:
Kreis , Mantelfläche , Kreisring
Berechne den (sichtbaren) Oberflächeninhalt des großen Zylinders (gr Z) und dann den (sichtbaren) Oberflächeninhalt des kleinen Zylinders (kl Z).
O g r Z \displaystyle O_{gr\; Z} O g r Z = = = Grundfl a ¨ che Kreis + Mantel + Kreisring \displaystyle \text{Grundfläche Kreis + Mantel + Kreisring} Grundfl a ¨ che Kreis + Mantel + Kreisring = = = r 1 2 ⋅ π + 2 ⋅ r 1 ⋅ π ⋅ h 1 + π ⋅ ( r 1 2 − r 2 2 ) \displaystyle r_1^2\cdot\pi+2\cdot r_1\cdot\pi\cdot h_1+\pi\cdot(r_1^2-r_2^2) r 1 2 ⋅ π + 2 ⋅ r 1 ⋅ π ⋅ h 1 + π ⋅ ( r 1 2 − r 2 2 ) = = = ( 3 cm ) 2 ⋅ π + 2 ⋅ 3 cm ⋅ π ⋅ 3 , 5 cm + π ⋅ ( ( 3 cm ) 2 − ( 1 cm ) 2 ) \displaystyle (3\;\text{cm})^2\cdot\pi+2\cdot 3\;\text{cm}\cdot\pi\cdot 3{,}5\;\text{cm}+\pi\cdot(( 3\;\text{cm})^2-(1\;\text{cm})^2) ( 3 cm ) 2 ⋅ π + 2 ⋅ 3 cm ⋅ π ⋅ 3 , 5 cm + π ⋅ (( 3 cm ) 2 − ( 1 cm ) 2 ) ↓
π ≈ 3 , 14 \pi \approx 3{,}14 π ≈ 3 , 14
≈ ≈ ≈ ( 3 cm ) 2 ⋅ 3 , 14 + 2 ⋅ 3 cm ⋅ 3 , 14 ⋅ 3 , 5 cm + 3 , 14 ⋅ ( ( 3 cm ) 2 − ( 1 cm ) 2 ) \displaystyle (3\;\text{cm})^2\cdot3{,}14+2\cdot 3\;\text{cm}\cdot3{,}14\cdot 3{,}5\;\text{cm}+3{,}14\cdot((3\;\text{cm})^2-(1\;\text{cm})^2) ( 3 cm ) 2 ⋅ 3 , 14 + 2 ⋅ 3 cm ⋅ 3 , 14 ⋅ 3 , 5 cm + 3 , 14 ⋅ (( 3 cm ) 2 − ( 1 cm ) 2 ) = = = 28 , 26 c m 2 + 65 , 94 c m 2 + 25 , 12 c m 2 \displaystyle 28{,}26\;\mathrm{cm^2}+65{,}94\;\mathrm{cm^2}+25{,}12\;\mathrm{cm^2} 28 , 26 c m 2 + 65 , 94 c m 2 + 25 , 12 c m 2 = = = 119 , 32 c m 2 \displaystyle 119{,}32\;\mathrm{cm^2} 119 , 32 c m 2
O k l Z \displaystyle O_{kl\; Z} O k l Z = = = Mantel + Deckfl a ¨ che Kreis \displaystyle \text{ Mantel + Deckfläche Kreis} Mantel + Deckfl a ¨ che Kreis = = = 2 ⋅ r 2 ⋅ π ⋅ h k + r 2 2 ⋅ π \displaystyle 2\cdot r_2\cdot\pi\cdot h_k+ r_2^2\cdot\pi 2 ⋅ r 2 ⋅ π ⋅ h k + r 2 2 ⋅ π = = = 2 ⋅ 1 cm ⋅ π ⋅ 4 cm + ( 1 cm ) 2 ⋅ π \displaystyle 2\cdot 1\;\text{cm}\cdot\pi\cdot 4\;\text{cm}+(1\;\text{cm})^2\cdot\pi 2 ⋅ 1 cm ⋅ π ⋅ 4 cm + ( 1 cm ) 2 ⋅ π ↓
π ≈ 3 , 14 \pi \approx 3{,}14 π ≈ 3 , 14
≈ ≈ ≈ 2 ⋅ 1 cm ⋅ 3 , 14 ⋅ 4 cm + ( 1 cm ) 2 ⋅ 3 , 14 \displaystyle 2\cdot 1\;\text{cm}\cdot3{,}14\cdot 4\;\text{cm}+(1\;\text{cm})^2\cdot3{,}14 2 ⋅ 1 cm ⋅ 3 , 14 ⋅ 4 cm + ( 1 cm ) 2 ⋅ 3 , 14 = = = 25 , 12 c m 2 + 3 , 14 c m 2 \displaystyle 25{,}12\;\mathrm{cm^2}+3{,}14\;\mathrm{cm^2} 25 , 12 c m 2 + 3 , 14 c m 2 = = = 28 , 26 c m 2 \displaystyle 28{,}26\;\mathrm{cm^2} 28 , 26 c m 2
O g e s a m t \displaystyle O_{gesamt} O g es am t = = = O g r Z + O k l Z \displaystyle O_{gr\;Z}+O_{kl\;Z} O g r Z + O k l Z = = = 119 , 32 c m 2 + 28 , 26 c m 2 \displaystyle 119{,}32\;\mathrm{cm^2}+28{,}26\;\mathrm{cm^2} 119 , 32 c m 2 + 28 , 26 c m 2 = = = 147 , 58 c m 2 \displaystyle 147{,}58\;\mathrm{cm^2} 147 , 58 c m 2
Antwort: Das Werkstück hat einen Oberflächeninhalt von etwa 147 , 58 c m 2 147{,}58 \;\mathrm{cm^2} 147 , 58 c m 2
Alternative Berechnung der Oberfläche
Die Oberfläche eines Zylinders berechnest du mit der Formel: O Z y l i n d e r = 2 ⋅ r 2 ⋅ π + 2 ⋅ r ⋅ π ⋅ h k O_{Zylinder} = 2\cdot r^2\cdot\pi+2\cdot r\cdot\pi\cdot h_k O Z y l in d er = 2 ⋅ r 2 ⋅ π + 2 ⋅ r ⋅ π ⋅ h k
Beachte : Die Fläche des g r u ¨ n e n \textcolor{006400}{grünen} g r u ¨ n e n zwei mal abgezogen werden, da diese Fläche sowohl beim großen als auch beim kleinen Zylinder nicht sichtbar ist.
O Z g e s a m t \displaystyle O_{Z\;gesamt} O Z g es am t = = = O g r Z + O k l Z − 2 ⋅ r 2 2 ⋅ π \displaystyle O_{gr\; Z}+O_{kl\;Z}-\textcolor{006400}{2 \cdot r_2^2\cdot \pi} O g r Z + O k l Z − 2 ⋅ r 2 2 ⋅ π = = = ( 2 ⋅ r 1 2 ⋅ π + 2 ⋅ r 1 ⋅ π ⋅ h 1 ) + ( 2 ⋅ r 2 2 ⋅ π + 2 ⋅ r 2 ⋅ π ⋅ h 2 ) − 2 ⋅ r 2 2 ⋅ π \displaystyle (2\cdot r_1^2\cdot\pi+2\cdot r_1\cdot\pi\cdot h_1)+(2\cdot r_2^2\cdot\pi+2\cdot r_2\cdot\pi\cdot h_2)-\textcolor{006400}{2\cdot r_2^2\cdot\pi} ( 2 ⋅ r 1 2 ⋅ π + 2 ⋅ r 1 ⋅ π ⋅ h 1 ) + ( 2 ⋅ r 2 2 ⋅ π + 2 ⋅ r 2 ⋅ π ⋅ h 2 ) − 2 ⋅ r 2 2 ⋅ π ↓
Die Fläche des kleinen (grünen) Kreises wurde zwei mal abgezogen.
= = = 2 ⋅ r 1 2 ⋅ π + 2 ⋅ r 1 ⋅ π ⋅ h 1 + 2 ⋅ r 2 ⋅ π ⋅ h 2 \displaystyle 2\cdot r_1^2\cdot\pi+2\cdot r_1\cdot\pi\cdot h_1+2\cdot r_2\cdot\pi\cdot h_2 2 ⋅ r 1 2 ⋅ π + 2 ⋅ r 1 ⋅ π ⋅ h 1 + 2 ⋅ r 2 ⋅ π ⋅ h 2 = = = 2 ⋅ ( 3 cm ) 2 ⋅ π + 2 ⋅ 3 cm ⋅ π ⋅ 3 , 5 cm + 2 ⋅ 1 cm ⋅ π ⋅ 4 cm \displaystyle
2\cdot (3\;\text{cm})^2\cdot\pi+2\cdot 3\;\text{cm}\cdot\pi\cdot 3{,}5\;\text{cm}+2\cdot 1\;\text{cm}\cdot\pi\cdot 4\;\text{cm} 2 ⋅ ( 3 cm ) 2 ⋅ π + 2 ⋅ 3 cm ⋅ π ⋅ 3 , 5 cm + 2 ⋅ 1 cm ⋅ π ⋅ 4 cm ↓
π ≈ 3 , 14 \pi \approx 3{,}14 π ≈ 3 , 14
≈ ≈ ≈ 2 ⋅ ( 3 cm ) 2 ⋅ 3 , 14 + 2 ⋅ 3 cm ⋅ 3 , 14 ⋅ 3 , 5 cm + 2 ⋅ 1 cm ⋅ 3 , 14 ⋅ 4 cm \displaystyle 2\cdot (3\;\text{cm})^2\cdot3{,}14+2\cdot 3\;\text{cm}\cdot3{,}14\cdot 3{,}5\;\text{cm}+2\cdot 1\;\text{cm}\cdot3{,}14\cdot 4\;\text{cm} 2 ⋅ ( 3 cm ) 2 ⋅ 3 , 14 + 2 ⋅ 3 cm ⋅ 3 , 14 ⋅ 3 , 5 cm + 2 ⋅ 1 cm ⋅ 3 , 14 ⋅ 4 cm ≈ ≈ ≈ 56 , 52 c m 2 + 65 , 94 c m 2 + 25 , 12 c m 2 \displaystyle
56{,}52\;\mathrm{cm^2}+65{,}94\;\mathrm{cm^2}+25{,}12\;\mathrm{cm^2} 56 , 52 c m 2 + 65 , 94 c m 2 + 25 , 12 c m 2 ≈ ≈ ≈ 147 , 58 c m 2 \displaystyle 147{,}58 \;\mathrm{cm^2} 147 , 58 c m 2
Antwort: Das Werkstück hat einen Oberflächeninhalt von etwa 147 , 58 c m 2 147{,}58 \;\mathrm{cm^2} 147 , 58 c m 2
