9.4 Volumes of Solids of Revolution: The Shell Method

Let \(R\) be the region under the curve \(y=f(x)\) between \(x=a\) and \(x=b\) (\(0\leq a<b\)) (Figure 1(a)). In Section 9.2, we computed the volume of the solid obtained by revolving \(R\) about the \(x\)-axis. Another way of generating a totally different solid is to revolve the region \(R\) about the \(y\)-axis as shown in Figure 1(b).

(a)

(b)

Figure 1

To compute the volume of this solid, consider a vertical thin rectangle of height \(f(x)\) and width \(dx\). When this rectangle is revolved around the \(y\)-axis, it generates a hollow, thin-walled shell of radius \(x\), height \(f(x)\) and thickness \(dx\) (Figure 2(a)).

If the cylindrical shell has been rolled out flat like a thin sheet of tin, it becomes a thin slab of height $f(x)$, thickness $dx$, and length \(2\pi x\), which is the circumference of the shell (Figure 2(b)). Therefore, the element of volume is \[dV=\underbrace{2\pi x}_{\small \rm circumference}\underbrace{f(x)}_{\small \rm height}\underbrace{dx}_{\small \rm thickness}.\] The total volume is then obtained by adding up the columns of the infinitesimal shells. \[V=\int_{a}^{b}dV=\int_{a}^{b}2\pi xf(x)dx.\]

(a)

(b)

Figure 2

In principle, the volume of this solid can also be obtained by considering thin disks generated by revolving infinitesimally thin horizontal rectangles; however, it often turns out to be more difficult because (1) the equation \(y=f(x)\) has to be solved for \(x\) in terms of \(y\), and (2) the formula for the length of the horizontal rectangle may vary in the region. In such cases, we have to compute more than one integral.

If the region is between two curves \(y=f(x)\) and \(y=g(x)\) (with \(f(x)\geq g(x)\)), then the height of the vertical rectangle, which is the same as the height of the cylindrical shell, is \(f(x)-g(x)\) (Figure 3). Therefore, in this case \[dV=2\pi x[f(x)-g(x)]dx\] and \[V=2\pi\int_{a}^{b}x[f(x)-g(x)]dx.\]

Figure 3

In general, we can write \[ \bbox[#F2F2F2,5px,border:2px solid black]{V=\int_{a}^{b}2\pi(\text{shell radius})(\text{area of thin rectangle})=\int_{a}^{b}2\pi\rho dA}\]

Example 1

The region in the first quadrant between \(y=4-x^{2}\) and \(y=x^{2}-4\) is revolved about the \(y\)-axis. Find the volume of the resulting solid by the shell method.

Solution 1

Figure 4

Let \(y_{1}=4-x^{2}\) and \(y_{2}=x^{2}-4\). Then the height of the typical shell \(MN\) is \[\begin{aligned} MN & =y_{1}-y_{2}\\ & =\left(4-x^{2}\right)-\left(x^{2}-4\right)\\ & =8-2x^{2}\end{aligned}\] So the volume of the shell shown in the figure is \[dV=2\pi\rho\ dA,\] where \[dA=MN\ dx=\left(8-2x^{2}\right)dx\] and \[\rho=\text{the distance of the thin rectangle from the axis of rotation}=x.\] Therefore, \[dV=2\pi x\left(8-2x^{2}\right)dx\] and the total volume is \[\begin{aligned} V & =\int_{0}^{2}dV=\int_{0}^{2}2\pi x\left(8-2x^{2}\right)dx\\ & =4\pi\int_{0}^{2}\left(4x-x^{3}\right)dx\\ & =4\pi\left[2x^{2}-\frac{1}{4}x^{4}\right]_{0}^{2}\\ & =4\pi\left(8-\frac{16}{4}\right)=16\pi.\end{aligned}\]

Example 2

The region inside the circle \(x^{2}+y^{2}=a^{2}\) is revolved about the \(y\)-axis. Find the volume of the resulting solid (which is a sphere) by the shell method.

Solution 2

Figure 5

We write the equation of the circle as \[y_{1}=\sqrt{a^{2}-x^{2}}\quad\text{and}\quad y_{2}=-\sqrt{a^{2}-x^{2}}.\] Then \[\begin{aligned} dA & =\text{area of the thin rectangle}\\ & =(y_{1}-y_{2})dx\\ & =2\sqrt{a^{2}-x^{2}}dx\end{aligned}\] and the volume of the shell is \[dV=2\pi\rho\,dA\] where \[\rho=\text{distance of the thin rectangle from the axis of rotation}=x.\] Therefore, \[dV=2\pi x\left(2\sqrt{a^{2}-x^{2}}\right)dx.\] Because \(x\) (= the distance of the thin rectangle from the \(y\)-axis) varies between \(0\) and \(a\), the total volume is: \[\begin{aligned} V & =\int_{0}^{a}dV\\ & =\int_{0}^{a}2\pi x\left(2\sqrt{a^{2}-x^{2}}\right)dx\\ & =2\pi\int_{0}^{a}2x\sqrt{a^{2}-x^{2}}dx\end{aligned}\] Let \(u=a^{2}-x^{2}\). Then \(du=-2x\,dx\) and \[\begin{aligned} V & =-2\pi\int_{-a^{2}}^{0}\sqrt{u}\,du\qquad\small{(u=0 \text{ when } x=a \text{ and } u=-a^{2} \text{ when } x=0)}\\ & =2\pi\int_{0}^{a^{2}}\sqrt{u}\,du\\ & =2\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{0}^{a^{2}}\\ & =\frac{4\pi}{3}a^{3}\end{aligned}\]

Example 3

The region inside the curve \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\] is revolved about the \(y\)-axis. Find the volume of the resulting ellipsoid.

Solution 3

Figure 6

If we solve the equation of the ellipse for \(y\), we get \[y_{1}=b\sqrt{1-\frac{x^{2}}{a^{2}}},\quad y_{2}=-b\sqrt{1-\frac{x^{2}}{a^{2}}}.\] Then \[\begin{aligned} dA & =\text{area of the thin rectangle}\\ & =(y_{1}-y_{2})dx\\ & =2b\sqrt{1-\frac{x^{2}}{a^{2}}}dx\end{aligned}\] and the volume of the shell is \[\begin{aligned} dV & =2\pi xdA\\ & =2\pi x\left(2b\sqrt{1-\frac{x^{2}}{a^{2}}}\right)\,dx\end{aligned}\] Because \(x\) (= the distance between the thin rectangle and the \(y\)-axis) varies between \(0\) and \(a\), the total volume is \[\begin{aligned} V & =\int_{0}^{a}dV\\ & =\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{a^{2}}}\,dx\end{aligned}\] Let \(u=1-\frac{x^{2}}{a^{2}}\). Then \[du=\frac{-2}{a^{2}}x\,dx.\] We know that \(u=1\) when \(x=0\) and \(u=0\) when \(x=a\). Therefore \[\begin{aligned} V & =\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{b^{2}}}dx\\ & =4b\pi\int_{1}^{0}\sqrt{u}\underbrace{\left(-\frac{a^{2}}{2}du\right)}_{xdx}\\ & =-2a^{2}b\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}\\ & =\frac{4}{3}\pi a^{2}b\end{aligned}\]

Example 4

Given that a disk with radius \(1\) and center \((5,0)\) is revolved about the \(y\)-axis, compute the volume of the resulting doughnut-shaped solid.

Solution 4

Figure 7

The equation of a circle with radius \(1\) and center \((5,0)\) is \[(x-5)^{2}+y^{2}=1\] This equation gives us two functions \[y_{1}=\sqrt{1-(x-5)^{2}}\quad\text{and}\quad y_{2}=-\sqrt{1-(x-5)^{2}}\] Therefore, the height of the typical thin rectangle is \[\begin{aligned} MN & =y_{1}-y_{2}\\ & =\sqrt{1-(x-5)^{2}}-\left(-1\sqrt{1-(x-5)^{2}}\right)\\ & =2\sqrt{1-(x-5)^{2}},\end{aligned}\] and the area of a thin rectangle is \[\begin{aligned} dA & =(MN)\,dx\\ & =2\sqrt{1-(x-5)^{2}}dx,\end{aligned}\] and the volume of the shell is \[\begin{aligned} dV & =2\pi\rho\,dA\\ & =2\pi x\left(2\sqrt{1-(x-5)^{2}}\right)dx\\ & =4\pi x\sqrt{1-(x-5)^{2}}dx.\end{aligned}\] Because \(\rho=x\) (\(=\) the distance between the thin rectangle and the origin) varies between \(4\) and \(6\), the total volume is \[V=\int_{4}^{6}dV=\int_{4}^{6}4\pi x\sqrt{1-(x-5)^{2}}dx\] Let \(u=x-5\). Then \[\begin{aligned} V & =\int_{-1}^{1}4\pi({\color{red}u}+{\color{blue}5})\sqrt{1-u^{2}}du\\ & =4\pi\int_{-1}^{1}{\color{red}u}\sqrt{1-u^{2}}du+4\pi\int_{-1}^{2}{\color{blue}5}\sqrt{1-u^{2}}du\end{aligned}\] Because the function \(f(u)=u\sqrt{1-u^{2}}\) is odd, that is \[f(-u)=-u\sqrt{1-(-u)^{2}}=-f(u)\] we know \[\int_{-1}^{1}u\sqrt{1-u^{2}}du=0\] On the other hand, \(\int_{-1}^{1}\sqrt{1-u^{2}}du\) is equal to the area of a semicircle of radius \(1\) (Figure 8). Therefore \[\int_{-1}^{1}\sqrt{1-u^{2}}du=\frac{\pi}{2}(1)^{2}=\frac{\pi}{2}\]

Therefore, the volume of the doughnut-shaped solid is \[\begin{aligned} V & =4\pi\underbrace{\int_{-1}^{1}u\sqrt{1-u^{2}}du}_{=0}+20\pi\underbrace{\int_{-1}^{1}\sqrt{1-u^{2}}du}_{=\frac{\pi}{2}}\\ & =10\pi^{2}.\end{aligned}\]

If a circle is revolved about an axis, the doughnut-shaped solid is called a torus. If the radius of the circle is \(r\) and the distance between the center of the circle and the axis is \(a\) then the volume of the torus is \[V=\left(\pi r^{2}\right)(2\pi a)=2\pi^{2}r^{2}a.\]