Find the volume of the solid of revolution generated by revolving the region bounded by the graphs of \(y=2x\) and \(y=x^{2}\) about the line \(x=2\).

Solution This example is similar to Example 5 except that the region is revolved about the line \(x=2\). Figure 27 shows the graph of the region, a typical washer, and the solid of revolution.

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Since the region is revolved about the vertical line \(x=2\), we express \(y=2x\) and \(y=x^2\) as \(x=\dfrac{y}{2}\) and \(x=\sqrt{y}\). The outer radius is \(2-\dfrac{y}{2}\) and the inner radius is \(2-\sqrt{y}\). The volume \(V\) of the solid of revolution is \begin{eqnarray*} V &=&\pi \displaystyle \int_{0}^{4}\left[ \left( 2-\dfrac{y}{2}\right) ^{2}-( 2- \sqrt{y}) ^{2}\right] ~{\it dy}\\[4pt] &=&\pi \displaystyle \int_{0}^{4}\left[ \left( 4-2y+\dfrac{ y^{2}}{4}\right) -( 4-4\sqrt{y}+y) \right] ~{\it dy} \\[4pt] &=&\pi \displaystyle \int_{0}^{4}\left( \dfrac{y^{2}}{4}-3y+4\sqrt{y}\right) ~{\it dy}=\pi \left[ \dfrac{y^{3}}{12}-\dfrac{3y^{2}}{2}+\dfrac{8y^{3/2}}{3}\right] _{0}^{4}=\dfrac{8}{3}\pi \hbox{ cubic units} \end{eqnarray*}