Use the disk method to find the volume of the solid of revolution generated by revolving the region bounded by the graph of $y=x^{3}$, the $y$-axis, and the lines $y=1$ and $y=8$ about the $y$-axis.

Solution Figure 20(a) shows the region to be revolved. Since the solid is formed by revolving the region about the $y$-axis, we write $y=x^{3}$ as $x=\sqrt[3]{y}=y^{1/3}$. Figure 20(b) illustrates a typical disk, and Figure 20(c) shows the solid of revolution. Using the disk method, the volume $V$ of the solid of revolution is $$\begin{eqnarray*} V&=&\pi \int_{1}^{8}[ y^{1/3}] ^{2}~{\it dy}=\pi \int_{1}^{8}y^{2/3}~{\it dy}=\pi \left[ \dfrac{y^{5/3}}{\dfrac{5}{3}}\right]_{1}^{8}\\[8pt] &=&\dfrac{3\pi }{5}(32-1)=\dfrac{93}{5}\pi \hbox{ cubic units} \end{eqnarray*}$$

Figure 20