Ratio of volumes

If the heights of two cylindrical drums are in the ratio 7:8 and their base radii are in the ratio 4:3. What is the ratio of their volumes?

Correct result:

p =  14:9

Solution:

p1=78=0.875 p2=43≐1.3333  p1=h1:h2 p2=r1:r2  V1=πr12 h1 V2=πr22 h2  V1=π⋅ (p2⋅ r2)2⋅ (p1⋅ h2) V1=π⋅ p22⋅ r22⋅ p1⋅ h2=p22⋅ p1⋅ V2   p=V1/V2  p=p22⋅ p1=1.33332⋅ 0.875=149≈1.5556=14:9p_{1}=\dfrac{ 7 }{ 8 }=0.875 \ \\ p_{2}=\dfrac{ 4 }{ 3 } \doteq 1.3333 \ \\ \ \\ p_{1}=h_{1}:h_{2} \ \\ p_{2}=r_{1}:r_{2} \ \\ \ \\ V_{1}=\pi r_{1}^2 \ h_{1} \ \\ V_{2}=\pi r_{2}^2 \ h_{2} \ \\ \ \\ V_{1}=\pi \cdot \ (p_{2} \cdot \ r_{2})^2 \cdot \ (p_{1} \cdot \ h_{2}) \ \\ V_{1}=\pi \cdot \ p_{2}^2 \cdot \ r_{2}^2 \cdot \ p_{1} \cdot \ h_{2}=p_{2}^2 \cdot \ p_{1} \cdot \ V_{2} \ \\ \ \\ \ \\ p=V_{1}/V_{2} \ \\ \ \\ p=p_{2}^2 \cdot \ p_{1}=1.3333^2 \cdot \ 0.875=\dfrac{ 14 }{ 9 }≈ 1.5556=14:9

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