Applications – Volumes of Revolution II

Problem 243: Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis.

(1) $\begin{equation*} y = x^2 + 1x - 42, \qquad y = 0, \end{equation*}$

about the $x$ -axis.

Solution: The volume of revolution is given by

(2) $\begin{equation*} V = \pi \int_{a}^{b}[R(x)]^2 \ dx,\end{equation*}$

where $R(x)$ is the distance between the two curves. We are not given the limits of integration. We need to find them. That is,

(3) $\begin{equation*}y = x^2 + 1x - 42 \implies 0 = x^2 + 1x - 42\implies x = -7, 6. \end{equation*}$

We now have

(4) $\begin{align*} V & = \pi \int_{a}^{b}[R(x)]^2 \ dx \\ & = \pi \int_{-7}^{6}[x^2+1x-42 ]^2 \ dx \\ & = \pi \frac{371293}{30} \\ & \approx 38881.71\end{align*}$

That is,

(5) $\begin{equation*} V = \frac{371293}{30} \pi. \end{equation*}$

One Problem at a Time

Applications – Volumes of Revolution II

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