Truncated cone

Calculate the volume of a truncated cone with base radiuses r1=13 cm, r2 = 10 cm and height v = 8 cm.

Result

V =  3342.7 cm3

Solution:

v2=r2r1−r2v=1013−108=26.67 V=V1−V2=13π(r12(v+v2)−r22v2)=3342.7 cm3 v_2 = \dfrac{r_2}{r_1-r_2}v = \dfrac{ 10}{ 13-10} 8 = 26.67 \ \\ V = V_1-V_2 = \dfrac{1}{3} \pi (r_1^2(v+v_2) - r_2^2v_2) = 3342.7 \ \text{cm}^3 \ \\

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