Truncated cone

Calculate the height of the rotating truncated cone with volume V = 794 cm³ and a base radii r₁ = 9.9 cm and r₂ = 9.8 cm.

Correct answer:

h =  2.6 cm

Step-by-step explanation:

$$\begin{gathered} r_1=9.9\text{ cm } \\ {r}_2=9.8\text{ cm } \\ V=794{\text{cm}}^3 \\ V={\displaystyle \frac{1}{3}}\pi h(r_1^2+r_1{r}_2+r_2^2) \\ h={\displaystyle \frac{3\cdot V}{\pi \cdot (r_1^2+r_1\cdot r_2+r_2^2)}}={\displaystyle \frac{3\cdot 794}{3.1416\cdot (9.9^2+9.9\cdot 9.8+9.8^2)}}=2.6\text{ cm} \end{gathered}$$

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Showing 5 comments:

Dr Math

it's formula; can be proved....https://math.stackexchange.com/questions/1626218/calculation-of-rise-in-height-of-water-in-a-frustum-of-right-circular-cone

2 years ago  Like

Math student

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

2 years ago  3 Likes Like

Dr Math

Fine math problem! Go ahead!

2 years ago  Like

Kukoslav

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

2 years ago  Like

Kukoslav

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

2 years ago  Like

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