Using the Washer Method: Revolving About a Horizontal Line

Find the volume \(V\) of the solid of revolution generated by revolving the region bounded by the graphs of \(y=2x\) and \(y=x^{2}\) about the line \(y=-5\).

Solution The region to be revolved is the same as in Example 5, but it is now being revolved about the line \(y=-5\). Figure 26 illustrates the region, a typical washer, and the solid of revolution.

Look at Figure 26(b). The outer radius of the washer is \(2u_i+5\) and the inner radius is \(u_i^{2}+5\). Partitioning the \(x\)-axis, the volume \(V_{i}\) of a typical washer is \[ V_{i}=\pi \Big[ ( 2u_{i}+5) ^{2}-\big( u_{i}^{2}+5\big) ^{2}\Big] \Delta x \]

The volume \(V\) of the solid of revolution as shown in Figure 26(c) is \begin{eqnarray*} V&=&\pi \displaystyle \int_{0}^{2}\Big[ ( 2x+5) ^{2}-( x^{2}+5) ^{2}\Big]\, {\it dx}=\pi \displaystyle \int_{0}^{2}( -x^{4}-6x^{2}+20x)\, {\it dx}\\[4pt] &=&\pi \left[ -\dfrac{x^{5}}{5}-2x^{3}+10x^{2}\right] _{0}^{2}=\dfrac{88}{5}\pi \hbox{ cubic units} \end{eqnarray*}