How to Use the Shell Method to Find a Volume of Revolution

The shell method finds the volume of a solid of revolution by adding the volumes of thin cylindrical shells. It is especially useful when slices perpendicular to the axis would give complicated washer or disk integrals.

The main rule is simple: take slices parallel to the axis of rotation. Each slice sweeps out a shell, and you integrate the shell volumes across the region.

The shell formula

A cylindrical shell with radius $r$, height $h$, and thickness $dx$ has approximate volume

$$\text{circumference} \times \text{height} \times \text{thickness} =2\pi r h\,dx.$$

Adding infinitely many shells gives

$$V=\int 2\pi\, (\text{radius})(\text{height})\, (\text{thickness}).$$

Use $dx$ for vertical slices and $dy$ for horizontal slices. The radius is the distance from a slice to the axis of rotation, while the height is the length of the slice inside the region.

Set up the integral

For a region rotated around a vertical line $x=c$, use vertical slices. A vertical slice at $x$ creates a shell with radius $|x-c|$ and height equal to the region’s vertical extent. If the region lies between $y=f(x)$ and $y=g(x)$, that height is $|f(x)-g(x)|$. Thus, when the region lies on one side of the axis,

$$V=2\pi\int_a^b |x-c|\,|f(x)-g(x)|\,dx.$$

For a region rotated around a horizontal line $y=c$, use horizontal slices. A slice at $y$ creates a shell with radius $|y-c|$ and height equal to the region’s horizontal extent. If its left and right boundaries are $x=L(y)$ and $x=R(y)$, then

$$V=2\pi\int_{y_0}^{y_1} |y-c|\,|R(y)-L(y)|\,dy.$$

Absolute values give distances and lengths. In many textbook problems the region stays entirely on one side of the axis, so the distances can be written without absolute values. If the region crosses the axis, split it where needed and check whether rotating separate parts produces overlapping shells; simply adding their volumes can count the same space twice.

Worked example: rotate around the y-axis

Find the volume formed by rotating the region under $y=x^2$, above $y=0$, and between $x=0$ and $x=2$ around the $y$-axis.

1. The axis is vertical, so use vertical slices, which are parallel to the axis.
2. At position $x$, the slice is $x$ units from the $y$-axis, so $r=x$.
3. Its height is the distance from $y=0$ to $y=x^2$, so $h=x^2$.
4. The slice positions run from $x=0$ to $x=2$.

Therefore,

$$\begin{aligned} V&=2\pi\int_0^2 x(x^2)\,dx\\ &=2\pi\int_0^2 x^3\,dx\\ &=2\pi\left[\frac{x^4}{4}\right]_0^2\\ &=8\pi. \end{aligned}$$

The volume is $8\pi$ cubic units. As a check, the washer method uses horizontal slices: at height $y$, the disk radius is $\sqrt{y}$, for $0\le y\le4$. It gives $\pi\int_0^4(\sqrt{y})^2\,dy=8\pi$ as well.

A shifted axis

The radius is not necessarily the coordinate itself. For example, rotating around $x=3$ makes a vertical slice at $x$ have radius $|3-x|$. For a region bounded by $y=0$, $y=x$, $x=0$, and $x=2$, its shell height is $x$, so

$$V=2\pi\int_0^2 (3-x)x\,dx=\frac{20\pi}{3}.$$

This illustrates why you should calculate radius as a distance from the axis, rather than automatically using $x$ or $y$.

Common mistakes

• Choosing slices perpendicular to the axis: those slices usually produce disks or washers. Shells come from slices parallel to the axis.
• Using the wrong radius: measure from the slice to the actual axis, including any shift such as $x=c$ or $y=c$.
• Using the wrong shell height: for vertical slices, height is a vertical distance; for horizontal slices, it is a horizontal distance.
• Mixing up the integration variable and bounds: vertical slices use $dx$ and $x$-bounds; horizontal slices use $dy$ and $y$-bounds. If the region’s boundary functions are given in the other variable, rewrite them in the variable required by slices parallel to the axis; if that is inconvenient, consider a different method such as washers.
• Ignoring changes in the region’s boundary: if the top, bottom, left, or right edge changes across the interval, split the integral at the transition points.
• Forgetting units: the integrand has units of area times a thickness, so the result is a volume in cubic units.