# Rotary bodies

The rotating cone and the rotary cylinder have the same volume 180 cm3 and the same height v = 15 cm. Which of these two bodies has a larger surface area?

Result

S1 =  199.533 cm2
S2 =  208.199 cm2

#### Solution:

$V=180 \ \\ h=15 \ \\ V=\pi r^2 \ h / 3 \ \\ r_{1}=\sqrt{ 3 \cdot \ V/\pi/h }=\sqrt{ 3 \cdot \ 180/3.1416/15 } \doteq 3.3851 \ \\ s=\sqrt{ h^2+r_{1}^2 }=\sqrt{ 15^2+3.3851^2 } \doteq 15.3772 \ \\ S_{1}=\pi \cdot \ r_{1}^2+\pi \cdot \ r_{1} \cdot \ s=3.1416 \cdot \ 3.3851^2+3.1416 \cdot \ 3.3851 \cdot \ 15.3772 \doteq 199.5326 \doteq 199.533 \ \text{cm}^2$
$V=\pi r^2 \ h \ \\ r_{2}=\sqrt{ V/\pi/h }=\sqrt{ 180/3.1416/15 } \doteq 1.9544 \ \\ S_{2}=2 \pi \cdot \ r_{2}^2+2 \pi \cdot \ r_{2} \cdot \ h=2 \cdot \ 3.1416 \cdot \ 1.9544^2+2 \cdot \ 3.1416 \cdot \ 1.9544 \cdot \ 15 \doteq 208.1988 \doteq 208.199 \ \text{cm}^2$

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