The radii of the flat circular faces of a bucket are x and 2x. If the height of the bucket is 3x, what is the capacity of the bucket? (Assume \pi=\frac{22}{7})

Explanation

A bucket forms the frustum of a cone. Volume V = \frac{1}{3}\pi h (R^2 + r^2 + Rr). Substituting the given parameters: V = \frac{1}{3} \times \frac{22}{7} \times 3x \times ((2x)^2 + x^2 + (2x)(x)) = \frac{22}{7} x (4x^2 + x^2 + 2x^2) = \frac{22}{7} x (7x^2) = 22x^3.

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