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.

422

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 $$\begin{eqnarray*} V &=&\pi \displaystyle \int_{0}^{4}\left[ \left( 2-\dfrac{y}{2}\right) ^{2}-( 2- \sqrt{y}) ^{2}\right] ~{\it dy}\\[4pt] &=&\pi \displaystyle \int_{0}^{4}\left[ \left( 4-2y+\dfrac{ y^{2}}{4}\right) -( 4-4\sqrt{y}+y) \right] ~{\it dy} \\[4pt] &=&\pi \displaystyle \int_{0}^{4}\left( \dfrac{y^{2}}{4}-3y+4\sqrt{y}\right) ~{\it dy}=\pi \left[ \dfrac{y^{3}}{12}-\dfrac{3y^{2}}{2}+\dfrac{8y^{3/2}}{3}\right] _{0}^{4}=\dfrac{8}{3}\pi \hbox{ cubic units} \end{eqnarray*}$$