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

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 \[V=\left(\pi r^{2}\right)(2\pi a)=2\pi^{2}r^{2}a.\]

*Figure 9: : A torus and its volume.
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