Question

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Answer 1

If it's a circle with an equilateral triangle cross-section, what shape is the solid?

Answer 2

A cone? I'm thinking since it's kinda looks like one that's what a cone is created from but, not 100% sure

Answer 3

18 sqrt(3)

Answer 4

I would create a function. \(A(x)\), that describes the area at any given x.

as the x-sect is a triangle, its area A is \(A = \frac{1}{2} bh\) where \(b = \sqrt{9- x^2} - (-\sqrt{9- x^2}) = 2 \sqrt{9- x^2}\).

as the x-sect is an **equilateral triangle**, we know that \(\tan 60^o = \dfrac{h}{b/2} = \sqrt{3} \implies h = \dfrac{\sqrt{3} b}{2}\) so \(A(x) = \frac{1}{4} \sqrt{3} b^2 = \frac{1}{4} \sqrt{3} ( 2\sqrt{9- x^2})^2 = \sqrt{3} (9 - x^2)\)

Volume \(V = \int_{-3}^3 A(x) ~ dx\) so you must integrate \(\sqrt{3} \int\limits_{-3}^3~ ~ (9 - x^2) ~ dx \)

i get twice your stated answer from wolfram so someone's made a typo, most likely me,.....but that's the method....


https://www.wolframalpha.com/input/?i=integrate++%5Csqrt%7B3%7D+(9+-+x%5E2)+from+x+%3D+-3+to+x+%3D+3