Truncated cone

Calculate the height of the rotating truncated cone with volume V = 1115 cm3 and a base radii r1 = 7.9 cm and r2 = 9.7 cm.

Result

h =  4.57 cm

Solution:

r1=7.9 cm r2=9.7 cm V=1115 cm3  V=13πh(r12+r1 r2+r22)  h=3⋅ Vπ⋅ (r12+r1⋅ r2+r22)=3⋅ 11153.1416⋅ (7.92+7.9⋅ 9.7+9.72)≐4.5672≐4.57 cmr_{1}=7.9 \ \text{cm} \ \\ r_{2}=9.7 \ \text{cm} \ \\ V=1115 \ \text{cm}^3 \ \\ \ \\ V=\dfrac{ 1 }{ 3 } \pi h (r_{1}^2+r_{1} \ r_{2}+r_{2}^2) \ \\ \ \\ h=\dfrac{ 3 \cdot \ V }{ \pi \cdot \ (r_{1}^2+r_{1} \cdot \ r_{2}+r_{2}^2) }=\dfrac{ 3 \cdot \ 1115 }{ 3.1416 \cdot \ (7.9^2+7.9 \cdot \ 9.7+9.7^2) } \doteq 4.5672 \doteq 4.57 \ \text{cm}

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Leave us a comment of this math problem and its solution (i.e. if it is still somewhat unclear...):

Showing 5 comments:
#
Dr Math

#
Math student
can u explain why you do (r2 + rxR + R2) in the first step

1 year ago  3 Likes
#
Dr Math
Fine math problem! Go ahead!

#
Kukoslav
but to prove formula, you need to know how to solve integral

#
Kukoslav
need to solve a cubic equation, as obtained above, to find rises in heights... integral

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