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=πr2 h/3 r1=3 V/π/h=3 180/3.1416/153.3851 s=h2+r12=152+3.3851215.3772 S1=π r12+π r1 s=3.1416 3.38512+3.1416 3.3851 15.3772199.5326199.533 cm2V=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=πr2 h r2=V/π/h=180/3.1416/151.9544 S2=2π r22+2π r2 h=2 3.1416 1.95442+2 3.1416 1.9544 15208.1988208.199 cm2V=\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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