Finding the Volume of a Solid: Revolving About the \(y\)-Axis

Find the volume \(V\) of the solid generated by revolving the region bounded by the graphs of \(f(x) =x^{2}+2x\), the \(x\)-axis, and the line \(x=1\) about the \(y\)-axis.

Solution Using the shell method: In the shell method, when a region is revolved about the \(y\)-axis, we partition the \(x\)-axis and use vertical shells. Figure 31(a) illustrates the region to be revolved and a typical rectangle of height \(f(u_{i})\) and thickness \(\Delta x\) that will become a shell with average radius \(u_{i}\) when it is revolved about the \(y\)-axis. The volume of a typical shell is \(V_{i}=2\pi\) (Average radius)(Height)(Thickness) \(=2\pi u_{i} f(u_{i})\Delta x\), as shown in Figure 31(b). Figure 31(c) illustrates the solid of revolution. The volume \(V\) of the solid of revolution is \begin{eqnarray*} V&=&2\pi \int_{0}^{1}xf(x) {\it dx}=2\pi \int_{0}^{1}[ x(x^{2}+2x) ] ~{\it dx}=2\pi \int_{0}^{1}(x^{3}+2x^{2})~{\it dx}\\[4pt] &=&2\pi\! \left[ \dfrac{x^{4}}{4}+ \dfrac{2x^{3}}{3}\right] _{0}^{1}=\dfrac{11\pi }{6}\hbox{ cubic units} \end{eqnarray*}

Using the washer method: Using the washer method, a revolution about the \(y\)-axis requires integration with respect to \(y\). This means we need to find the inverse function of \(y=f(x)\). We treat \(x^{2}+2x-y=0\) as a quadratic equation in the variable \(x\) and use the quadratic formula with \(a=1, b=2\), and \(c=-y\) to obtain \(x=g(y)=-1\pm \sqrt{1+y}\). Since \(x\geq 0\), we use the \(+\) sign.

See Figure 32. The volume of a typical washer is \[ V_{i}=\pi [ (\hbox{Outer radius}) ^{2}-( \hbox{Inner radius}) ^{2}]\times (\hbox{Thickness}). \]

The volume \(V\) of the solid of revolution is \begin{eqnarray*} V &=&\pi \int_{0}^{3} [ 1^{2}- ( -1+\sqrt{1+y}) ^{2}] ~{\it dy}=\pi \int_{0}^{3} [ 1- ( 1-2\sqrt{1+y}+1+y) ] ~{\it dy}\\[4pt] &=&\pi \int_{0}^{3} [ 2\sqrt{1+y}-1-y]\, {\it dy} =\pi \int_{0}^{3} ( 2\sqrt{1+y}) {\it dy}-\pi \int_{0}^{3}(1+y)~{\it dy} \end{eqnarray*}

*This topic is discussed in detail in an article by Charles A. Cable (February 1984), “The Disk and Shell Method,” American Mathematical Monthly, 91(2), 139.

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The two integrals are found as follows:

\( \pi \displaystyle \int_{0}^{3}(2\sqrt{1+y})~ {\it dy} \underset{\underset{\underset{\color{#0066A7}{\hbox{then \({du}={dy}\)}}}{\color{#0066A7}{\hbox{Let \(u=1+y\);}}}}{\color{#0066A7}{\uparrow }}} {=} 2\pi \displaystyle \int_{1}^{4}u^{1/2}\ {\it du}=2\pi \left[ \dfrac{u^{3/2}}{\dfrac{3}{2}}\right]_{1}^{4}=\dfrac{28\pi }{3} \)

\(\pi \displaystyle \int_{0}^{3}(1+y)~{\it dy}=\pi \left[ y+\dfrac{y^{2}}{2}\right] _{0}^{3}= \dfrac{15\pi }{2}\)

The volume \(V\) is \[ V=\frac{28\pi }{3}-\frac{15\pi }{2}=\frac{11\pi }{6}\hbox{ cubic units} \]