The shell method slices a solid of revolution into thin cylindrical shells, like the layers of an onion or the rings of a tree trunk. For a region under \( y = f(x) \), \( a \le x \le b \) (with \( a \ge 0 \)), rotated about the y-axis:
$$V = 2\pi\int_a^b x\,f(x)\,dx$$
The big advantage: you integrate along the axis perpendicular to the axis of rotation, so you can keep the function in the form it was given. This guide shows where the formula comes from, how to set up shells about any vertical or horizontal line, and how to decide between shells and disks.
Where the formula comes from
A shell at position \( x \) with thickness \( dx \) has radius \( x \) and height \( f(x) \). Unroll it and you get a thin flat slab:
$$\text{circumference}\times\text{height}\times\text{thickness} = 2\pi x\cdot f(x)\cdot dx$$
Adding all the shells gives the integral. The general pattern is easy to remember:
$$V = 2\pi\int(\text{radius})(\text{height})\,dx$$
A more careful derivation
The unrolling picture is an approximation, so here is why it gives the exact answer.
Step 1: one exact shell. Split \( [a, b] \) into subintervals of width \( \Delta x \). Over \( [x_{i-1}, x_i] \), take a rectangle of height \( f(\bar x_i) \), where \( \bar x_i \) is the midpoint. Rotated about the y-axis, it becomes a hollow cylinder: an outer cylinder of radius \( x_i \) minus an inner one of radius \( x_{i-1} \).
Step 2: compute its volume. Using the difference of squares,
$$\begin{aligned} V_i &= \pi x_i^2 f(\bar x_i) - \pi x_{i-1}^2 f(\bar x_i) \\ &= \pi(x_i + x_{i-1})(x_i - x_{i-1})\,f(\bar x_i) \\ &= 2\pi\,\bar x_i\,f(\bar x_i)\,\Delta x \end{aligned}$$
because \( x_i + x_{i-1} = 2\bar x_i \). No approximation was needed in this step.
Step 3: add and take the limit. The total is a Riemann sum for \( 2\pi x f(x) \). As \( \Delta x \to 0 \), the stack of shells fills the solid and the sum becomes \( 2\pi\int_a^b x f(x)\,dx \).
Sanity check: a plain cylinder
Rotate the rectangle under \( y = 5 \), \( 0 \le x \le 2 \), about the y-axis. You should get a cylinder of radius 2 and height 5:
$$V = 2\pi\int_0^2 5x\,dx = 2\pi\cdot 10 = 20\pi = \pi(2^2)(5)$$
How to set up a shell integral
- Sketch the region and the axis of rotation.
- Draw a thin strip parallel to the axis. Rotating it makes one shell.
- The radius is the distance from the strip to the axis.
- The height is the length of the strip: top minus bottom (or right minus left).
- Find the limits from the region’s edges or intersection points, then integrate \( 2\pi(\text{radius})(\text{height}) \).
Example 1
Rotate the region under \( y = x - x^2 \), \( 0\le x\le1 \), about the y-axis.
$$V = 2\pi\int_0^1x(x - x^2)\,dx = 2\pi\left(\frac13 - \frac14\right) = \frac{\pi}{6}$$
With disks you’d have to solve \( y = x - x^2 \) for \( x \), giving two messy branches. Shells avoid that completely.
This is also the region between \( y = x \) and \( y = x^2 \), since the height \( x - x^2 \) is top minus bottom. The washer method with \( dy \) gives the same \( \frac\pi6 \).
Example 2
\( y = \sqrt x \), \( 0\le x\le4 \), about the y-axis:
$$V = 2\pi\int_0^4x^{3/2}\,dx = 2\pi\cdot\frac25\cdot32 = \frac{128\pi}{5}$$
Example 3
\( y = 4 - x^2 \), \( 0\le x\le2 \), about the y-axis:
$$V = 2\pi\int_0^2(4x - x^3)\,dx = 2\pi(8 - 4) = 8\pi$$
Example 4: needs integration by parts
\( y = \sin x \), \( 0\le x\le\pi \), about the y-axis:
$$V = 2\pi\int_0^\pi x\sin x\,dx = 2\pi\cdot\pi = 2\pi^2$$
(\( \int_0^\pi x\sin x\,dx = \pi \) by integration by parts: with \( u = x \) and \( dv = \sin x\,dx \), the antiderivative is \( \sin x - x\cos x \), which equals \( \pi \) at \( \pi \) and \( 0 \) at \( 0 \).) Disks would be much harder here, because you would need \( x = \arcsin y \) on two separate branches.
Rotating about a vertical line x = k
The radius becomes the distance to the axis. For the region of Example 1 rotated about \( x = 3 \), the radius is \( 3 - x \):
$$V = 2\pi\int_0^1(3 - x)(x - x^2)\,dx = \frac{5\pi}{6}$$
Example 5: an axis to the left
Rotate the region under \( y = x^2 \), \( 0 \le x \le 1 \), about \( x = -1 \). Now the axis is to the left of the region, so the radius is \( x - (-1) = x + 1 \). The height is still \( x^2 \):
$$\begin{aligned} V &= 2\pi\int_0^1(x + 1)x^2\,dx \\ &= 2\pi\left(\frac14 + \frac13\right) = \frac{7\pi}{6} \end{aligned}$$
The rule: if the axis is at \( x = k \), the radius is \( k - x \) when the axis is to the right and \( x - k \) when it is to the left.
Shells about the x-axis
For a horizontal axis, use horizontal strips and integrate in \( y \): \( V = 2\pi\int_c^d y\,h(y)\,dy \).
Example 6: checking a disk answer
Rotate the region under \( y = \sqrt x \), \( 0 \le x \le 4 \), about the x-axis. A horizontal strip at height \( y \) runs from the curve \( x = y^2 \) to the line \( x = 4 \), so its length is \( 4 - y^2 \), and \( y \) goes from 0 to 2:
$$V = 2\pi\int_0^2 y(4 - y^2)\,dy = 2\pi(8 - 4) = 8\pi$$
The disk method gives \( \pi\int_0^4 x\,dx = 8\pi \) as well. Here disks were easier, which shows why it pays to know both.
Disk or shell? A quick guide
| Axis of rotation | Your function is \( y = f(x) \) | Your function is \( x = g(y) \) |
|---|---|---|
| x-axis (horizontal) | disks/washers, \( dx \) | shells, \( dy \) |
| y-axis (vertical) | shells, \( dx \) | disks/washers, \( dy \) |
Rule of thumb: shells are parallel to the axis, disks are perpendicular to it. Pick whichever lets you integrate in the variable your function is already written in.
Two more tie-breakers: prefer shells when the region’s top or bottom boundary changes formula partway across (disks would need several integrals), and prefer disks when the solid has a hole that is easy to describe with an inner radius.
Common mistakes
- Dropping the 2π. The circumference of a shell is \( 2\pi r \), so the constant is \( 2\pi \), not \( \pi \).
- Using x as the radius for every axis. The radius is \( x \) only for the y-axis. For \( x = k \), use the distance \( |k - x| \).
- Wrong height between two curves. The height is top minus bottom, not just the top function.
- Wrong variable for a horizontal axis. Shells about the x-axis need \( dy \), with the height written as right minus left in terms of \( y \).
- Limits that cross the axis. The shell formula assumes the region stays on one side of the axis. If it straddles it, split the region or reconsider the setup.
Where it’s used
Shells turn up whenever a solid is naturally built from layers around a central axis: pipes, tree trunks, pressure vessels and rotating parts. The integrals often lead to integration by parts or a u-substitution, as in the practice problem with \( x e^{-x^2} \). In multivariable calculus, the factor \( 2\pi r \) reappears as the \( r \) in polar and cylindrical coordinates, so shells are good preparation for double integrals.
Shell method calculator
The full shell method calculator shows the integral and the volume side by side, so you can compare each piece with your own work.
Practice problems
Try these before checking the answers.
- \( y = x^2,\ 0\le x\le 2 \text{ about the y-axis} \)
- \( y = e^{-x^2},\ 0\le x\le 1 \text{ about the y-axis} \)
- \( y = \sqrt x,\ 0 \le x\le 1 \text{ about } x = 2 \)
- \( y = \frac1x,\ 1 \le x \le 3 \), about the y-axis
- \( y = x^2,\ 0 \le x \le 1 \), about \( x = 1 \)
- The region under \( y = \sqrt x \), \( 0 \le x \le 1 \), about the x-axis, using shells in \( y \)
Answers: (1) \( 8\pi \); (2) \( \pi\left(1 - \frac1e\right) \); (3) \( \frac{28\pi}{15} \); (4) \( 4\pi \); (5) \( \frac\pi6 \); (6) \( \frac\pi2 \), from \( 2\pi\int_0^1 y(1 - y^2)\,dy \).
FAQ
Do disks and shells give the same answer?
Always, if both are set up correctly. They’re two ways of slicing the same solid, and a great way to check your work.
What if the region is between two curves?
The height becomes top minus bottom: \( V = 2\pi\int x\,\big(f(x) - g(x)\big)\,dx \).
Why is the shell formula 2π and the disk formula π?
A disk’s volume uses the area \( \pi r^2 \) of a circle. A shell’s volume uses the circumference \( 2\pi r \) of a circle, multiplied by height and thickness. Different shapes, different constants.
When is the shell method easier than the disk method?
When the axis is vertical and the curve is given as \( y = f(x) \), or when solving for the other variable would be hard or impossible, as with \( y = \sin x \) or \( y = x - x^2 \) about the y-axis.
Can I use the shell method around the x-axis?
Yes. Use horizontal strips, radius \( y \) (or the distance to a horizontal line \( y = k \)), height written in terms of \( y \), and integrate with respect to \( y \).
Further reading
- Volumes of Solids of Revolution / Method of Cylinders (Paul’s Online Math Notes) — four worked examples with rotation about vertical, horizontal and shifted axes
- Volumes of Revolution: Cylindrical Shells (OpenStax Calculus Volume 1) — the textbook derivation, examples and a comparison with disks and washers
Calculators for this topic
Keep learning
Optimization Problems in Calculus: 4 Solved Examples
Solve calculus optimization problems in 6 steps. Worked examples: the largest fenced area, the open box, the cheapest can and the closest point on a curve.
Related Rates: Step-by-Step Strategy and Classic Problems
Solve related rates problems in 6 steps. Worked classics: the sliding ladder, the expanding balloon, the draining cone and the moving shadow.