OBJECTIVES

When you finish this section, you should be able to:

1. Use the shell method to find the volume of a solid formed by revolving a region about the \(y\)-axis (p. 425)
2. Use the shell method to find the volume of a solid formed by revolving a region about the \(x\)-axis (p. 428)
3. Use the shell method to find the volume of a solid formed by revolving a region about a line parallel to a coordinate axis (p. 430)

NOTE

Sewer lines and the water pipes in a house are cylindrical shells.

Volume of a Solid of Revolution About the \(y\)-Axis: the Shell Method

If \(y=f(x)\) is a function that is continuous and nonnegative on the closed interval \([a,b]\), where \(a\geq 0\), then the volume \(V\) of the solid generated by revolving the region bounded by the graph of \(f\), the \(x\)-axis, and the lines \(x = a\) and \(x = b\) about the \(y\)-axis is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ V=2\pi \int_{a}^{b}x f(x)~{\it dx}} \end{equation*}

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*}

Figure 31 The shell method.

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*}

Figure 32 The washer method.

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 \[ V=\frac{28\pi }{3}-\frac{15\pi }{2}=\frac{11\pi }{6}\hbox{ cubic units} \]

NOW WORK

Problem 5.

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 \(f(x) =x^{2}\) and \(g(x) =12-x\) to the right of \(x=1\) about the \(y\)-axis.

Solution Using the shell method: Figure 33(a) shows the graph of the region to be revolved and a typical rectangle.

Figure 33 The shell method.

As shown in Figure 33(b), in the shell method, we partition the \(x\)-axis and use vertical shells. A typical shell has height \(h_{i}=g( u_{i}) -f( u_{i}) =(12-u_{i})-u_{i}^{2}=12-u_{i}-u_{i}^{2}\), and volume \(V_{i}=2\pi u_{i}( 12-u_{i}-u_{i}^{2}) \Delta x\). Figure 33(c) shows the solid of revolution. Notice that the integration takes place from \(x=1\) to \(x=3\). The volume \(V\) of the solid of revolution is \begin{eqnarray*}\hskip-1pc V &=&2\pi \int_{1}^{3}x(12-x-x^{2})~{\it dx}=2\pi \int_{1}^{3}( 12x-x^{2}-x^{3}) ~{\it dx}=2\pi \left[ 6x^{2}-\dfrac{x^{3}}{3}-\dfrac{x^{4}}{ 4}\right] _{1}^{3} \\[4pt] &=&2\pi\! \left[ \left( 54-9-\dfrac{81}{4}\right) -\left( 6-\dfrac{1}{3}- \dfrac{1}{4}\right) \right] =\dfrac{116~\pi }{3}\hbox{ cubic units} \end{eqnarray*}

Using the washer method: Figure 34 shows the solid of revolution and typical washers.

Figure 34 The washer method.

In the washer method, we partition the interval \(\left[ 1,11\right]\) on the \(y\)-axis and use horizontal washers. At \(y=9\), the function on the right changes. The volume of a typical washer in the interval \([1,9]\) is \begin{equation*} V_{i}=\pi \big[ \sqrt{v_{i}}^{2}-1^{2}\big] \Delta y=\pi (v_{i}-1) \Delta y \end{equation*}

The volume of a typical washer in the interval \([9,11]\) is \begin{equation*} V_{i}=\pi \big[ ( 12-v_{i}) ^{2}-1^{2}\big] \Delta y=\pi \big( 143-24v_{i}+v_{i}^{2}\big) \Delta y \end{equation*}

The volume \(V\) of the solid of revolution is \begin{eqnarray*} V &=&\pi \int_{1}^{9}(y-1) ~{\it dy}+\pi \int_{9}^{11} (143-24y+y^{2})~{\it dy}\\[8pt] &=&\pi \left[ \dfrac{y^{2}}{2}-y\right] _{1}^{9}+\pi \left[ 143y-12y^2 +\dfrac{y^{3}}{3} \right] _{9}^{11}\hbox{ } \notag \\[8pt] &=&\pi \left[ \left( \dfrac{81}{2}-9\right) -\left( \dfrac{1}{2}-1\right) \right] +\pi \left( 143\left( 2\right) -(12)(121-81)+\dfrac{11^{3}}{3}-\dfrac{9^{3}}{3}\right)\\[8pt] &=&32\pi +\dfrac{20}{3}\pi =\dfrac{116\,\pi }{3}\hbox{ cubic units} \end{eqnarray*}

NOW WORK

Problem 29.

Volume of a Solid of Revolution About the \(x\)-Axis: the Shell Method

If \(x=g(y)\) is a function that is continuous and nonnegative on the closed interval \([c,d], c\geq 0\), the volume \(V\) of the solid generated by revolving the region bounded by the graphs of \(x=g(y)\), the \(y\)-axis, and the lines \(y=c\) and \(y=d\) about the \(x\)-axis is \begin{equation*} \bbox[5px, border:1px solid black, #F9F7ED]{ V=2\pi \int_{c}^{d}y~ g(y)~{\it dy}} \end{equation*}

Using the Shell Method: Revolving About the \(x\)-Axis

Find the volume \(V\) of the solid generated by revolving the region bounded by the graph of, \(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1, a > 0, b > 0\), in the first quadrant, about the \(x\)-axis.

Solution The equation \(\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1\) defines an ellipse. The intercepts of its graph are \((a,0), (0,b), (-a,0)\), and \((0,-b)\). The region to be revolved is the shaded region in the first quadrant shown in Figure 36(a).

Figure 36 The shell method.

In the shell method, when the region is revolved about the \(x\)-axis, we partition the \(y\)-axis and use horizontal shells. A revolution about the \(x\)-axis requires integration with respect to \(y\), so we express the equation of the ellipse as \[ x=g(y)=\dfrac{a}{b}\sqrt{~b^{2}-y^{2}} \]

The volume of a typical shell is \[ V_{i}=2\pi\ \hbox{(Average radius)(Height)(Thickness)} = 2\pi v_{i}g(v_{i})\Delta y \]

See Figure 36(b).

Figure 36(c) shows the solid of revolution. The volume \(V\) of the solid of revolution is \begin{eqnarray*} V &=&2\pi \int_{0}^{b}y~g(y)~{\it dy}=2\pi \int_{0}^{b}y\left( \dfrac{a}{b}\sqrt{ b^{2}-y^{2}}\right) {\it dy} \underset{\underset{\underset{\color{#0066A7}{\hbox{then \({du}={-2y}{ dy}\)}}}{\color{#0066A7}{\hbox{Let \(u=b^2-y^2\);}}}}{\color{#0066A7}{\uparrow }}} {=} 2\pi \dfrac{a}{b}\left( -\dfrac{1}{2}\right) \int_{b^{2}}^{0}\sqrt{u}\ {\it du} \\[-9pt] &=&\dfrac{\pi a}{b}\int_{0}^{b^{2}}u^{1/2}~{\it du}=\dfrac{\pi a}{b}\left[ \dfrac{ u^{3/2}}{\dfrac{3}{2}}\right] _{0}^{b^{2}}=\dfrac{2\pi a}{3b}( b^{3}) =\dfrac{2\pi ab^{2}}{3}\hbox{ cubic units} \end{eqnarray*}

NOW WORK

Problem 33.

Using the Shell Method: Revolving About the Line \(x=2\)

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

Solution The region bounded by the graph of \(y=2x-2x^{2}\) and the \(x\)-axis is illustrated in Figure 37(a). A typical shell formed by revolving the region about the line \(x=2\), as shown in Figure 37(b), has an average radius of \(2-u_{i}\), height \(f(u_{i}) = 2u_{i}-2u_{i}^{2}\), and thickness \(\Delta x\). The solid of revolution is depicted in Figure 37(c).

Figure 37

The volume \(V\) of the solid is \begin{eqnarray*} V&=&2\pi \displaystyle \int_{0}^{1}(2-x)(2x-2x^{2})~{\it dx}=4\pi \displaystyle \int_{0}^{1}(x^{3}-3x^{2}+2x)~{\it dx}\\[4pt] &=&4\pi \left[ \dfrac{x^{4}}{4}-x^{3}+x^{2} \right] _{0}^{1}=\pi \hbox{ cubic units} \end{eqnarray*}

NOW WORK

Problem 17.