Find the value of a function given the volume of a shape defined by its ordinate set

Consider a solid shape whose base is defined as the ordinate set of a function $f$ on an interval $[0,a]$ . The cross sections of this solid taken perpendicular to the interval are squares. The volume of the solid is given by the formula

$V = a^3 - 2a \cos a + (2-a^2) \sin a \qquad \text{for all } a \geq 0.$

If $f$ is continuous on $[0,a]$ calculate the value $f(a)$ .

We are given the expression

$V = a^3 - 2a \cos a + (2-a^2) \sin a.$

However, we also know we can compute the volume as the integral from 0 to $a$ of the area of each cross section. Since the cross sections are squares and the length of the edge on the base is given by $f(x)$ (since the base is the ordinate set of $f$ , the length of the base of a cross section at a point $x$ is $f(x) - 0 = f(x)$ ), we know the area of each cross section is $(f(x))^2$ . Therefore, we have another expression for the volume of the solid given by

$V = \int_0^a (f(x))^2 \, dx.$

Setting these two expressions for the volume equal and differentiating both sides we have

$\begin{align*} &&a^3 - 2a \cos a + (2-a^2) \sin a &= \int_0^a (f(x))^2 \, dx \\ \implies && 3a^2 - a^2 \cos a &= (f(a))^2 \\ \implies && f(a) &= a(3- \cos a)^{\frac{1}{2}}. \end{align*}$

Stumbling Robot

Find the value of a function given the volume of a shape defined by its ordinate set

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