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Find an interval on which the solid of revolution has a particular volume

Let

    \[ f(x) = \sqrt{x}, \qquad g(x) = \frac{x}{2}, \qquad 0 \leq x \leq 2.\]

Find a real number t \in [1,2] such that the solid of revolution generated by the region between f and g on the interval [0,t] has volume \frac{\pi t^3}{3}.


The volume of the solid of revolution generated by the region between the graphs of f and g on the interval [0,t] is,

    \[ V = \pi \int_0^t \left( x - \frac{x^2}{4} \right) \, dx. \]

So, we set this equal to \frac{\pi t^3}{3} and solve for t,

    \begin{align*}  && \frac{\pi t^3}{3} &= \pi \int_0^t \left(x - \frac{x^2}{4} \right) \, dt \\ \implies && \frac{t^3}{3} &= \left. \frac{x^2}{2} \right|_0^t - \left. \frac{x^3}{12} \right|_0^t \\ \implies && \frac{t^3}{3} &= \frac{t^2}{2} - \frac{t^3}{12} \\ \implies && 4t^3 &= 6t^2 - t^3 & (1 < t < 2 \implies t \neq 0)\\ \implies && t &= \frac{6}{5}. \end{align*}

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