Short and Sweet Calculus

7.4 Volumes: 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 [fig:Ch9-Shell-Fig1](a)). In Section [sec:Ch9-disk-method], 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 [fig:Ch9-Shell-Fig1](b).

(a) (b)

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 [fig:Ch9-Shell-Fig2](a)).

Imagine that the cylindrical shell has been rolled out flat like a thin sheet of tin whose length is \(2\pi x\), which is the circumference of the shell (Figure [fig:Ch9-Shell-Fig2](b)). Therefore, the element of volume is \[dV=2\pi xf(x)dx.\] 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)
  • 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 and the cylindrical shell are \(f(x)-g(x)\) (Figure 7.14). Therefore, in this case \[dV=2\pi x[f(x)-g(x)]dx\] and \[V=2\pi\int_{a}^{b}x[f(x)-g(x)]dx.\]

In general, we can write \[\boxed{V=\int_{a}^{b}2\pi(\text{shell radius})(\text{area of thin rectangle})=\int_{a}^{b}2\pi\rho dA}\]

Example 7.6. 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.


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 \[\begin{aligned} dA & =\text{volume of the thin rectangle}\\ & =(y_{1}-y_{2})dx\\ & =2b\sqrt{1-\frac{x^{2}}{a^{2}}}dx\end{aligned}\] and the volume of the shell is \[dV=2\pi xdA=2\pi x\left(2b\sqrt{1-\frac{x^{2}}{a^{2}}}\right)\,dx\] Because \(x\) (= the distance between the thin rectangle and the \(y\)-axis) varies between \(0\) and \(a\), the total volume is \[V=\int_{0}^{a}dV=\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{a^{2}}}\,dx\] 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 \[V=\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{b^{2}}}dx=4b\pi\int_{1}^{0}\sqrt{u}\underbrace{\left(-2\frac{a^{2}}{2}du\right)}_{xdx}=-2a^{2}b\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}=\frac{4}{3}\pi a^{2}b\]

[up][previous][table of contents][next]