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Find a function given the volume of a solid determined by the function

by RoRi

Given a solid with base defined by the ordinate set of a continuous function f on the interval [1,a]. The cross sections take perpendicular to [1,a] are in the shape of squares. Find the function f(a) if the volume of the solid is

    \[ \frac{1}{3} a^3 \log^2 a - \frac{2}{9} a^3 \log a + \frac{2}{27} a^3 - \frac{2}{27} \qquad \text{for all } a \geq 1. \]

The volume of the solid is equal to the integral,

    \[ V = \int_1^a (f(x))^2 \, dx \]

since the cross sections are in the shape of squares and the length of the base is f(x). So the cross sectional area at each point from 1 to a is (f(x))^2, and then the volume is obtained by integrating these areas over [1,a]. Setting this equal to the given formula for the volume,

    \begin{align*} &&\int_1^a (f(x))^2 \, dx &= \frac{1}{3}a^3 \log^2 a - \frac{2}{9} a^2 \log a + \frac{2}{27} a^3 - \frac{2}{27} \\ \implies && \frac{dV}{da} = (f(a))^2 &= a^2 \log^2 a + \frac{2}{3}a^2 \log a - \frac{2}{3} a^2 \log a - \frac{2}{9}a^2 + \frac{2}{9}a^2 \\ &&&= a^2 \log^2 a \\[10pt] \implies && f(a) &= a \log a. \end{align*}

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