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.

Solution

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 $V=\int_{0}^{a}4b\pi x\sqrt{1-\frac{x^{2}}{b^{2}}}dx=4b\pi\int_{1}^{0}\sqrt{u}\underbrace{\left(-2\frac{a^{2}}{2}du\right)}_{xdx}=-2a^{2}b\pi\left[\frac{2}{3}u^{\frac{3}{2}}\right]_{1}^{0}=\frac{4}{3}\pi a^{2}b$

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