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A Complete Shell-Method Volume Workshop

Learn a reliable workflow for setting up and evaluating cylindrical-shell integrals for bounded regions.

Intermediate 11 min read

What You'll Learn

  • Choose vertical or horizontal slices according to the axis of rotation.
  • Identify shell radius, shell height, and integration bounds from a region.
  • Apply the cylindrical-shell volume formula with the correct differential.
  • Evaluate exact volume integrals and check their antiderivatives.
  • Diagnose a common error involving distance from a shifted axis.

A Reliable Workflow for Shell Volumes

A shell-method problem is not just an integration exercise. The integral records a geometric decision: which thin slices of a region become cylindrical shells when the region is rotated? A correct setup depends on matching the slice orientation to the axis, measuring each radius as a distance from that axis, and finding the height and bounds from the region.

This workshop brings those choices together in three bounded-region problems. The examples use both vertical and horizontal slices, and one rotates around a shifted axis. In each case, we will describe the region, identify a representative slice, construct the integral, evaluate it, and check the calculation.

Definition: If a thin slice parallel to the axis of rotation sweeps out a cylindrical shell of radius \(r\), height \(h\), and thickness \(d\), its volume contribution is approximately \(2\pi rh\,d\). The shell method adds these contributions across the region: $$ V=2\pi\int r\,h\,d. $$ The variable of integration is the coordinate that measures slice thickness.

How the Shell Formula Arises

Consider a shell with fixed height \(h\), inner radius \(r\), and small thickness \(\Delta r\). It can be viewed as the difference between two cylinders. The outer cylinder has radius \(r+\Delta r\), so the shell volume is

$$ \pi(r+\Delta r)^2h-\pi r^2h = \pi\bigl(2r\Delta r+(\Delta r)^2\bigr)h. $$

After dividing by the thickness, the volume per unit thickness is

$$ \frac{\pi\bigl(2r\Delta r+(\Delta r)^2\bigr)h}{\Delta r} = 2\pi rh+\pi h\Delta r. $$

As the thickness tends to zero, the last term tends to zero. The contribution per unit thickness therefore tends to \(2\pi rh\). When the radius or height varies across the region, adding the thin-shell contributions and taking the limit gives the integral formula. This is the same limiting idea expressed by a Riemann integral.

Key decision: A vertical slice has thickness \(dx\); a horizontal slice has thickness \(dy\). For the usual shell setup, choose slices parallel to the axis of rotation. The radius is the distance from the slice to the axis, and the height is the length of the slice inside the region.

Worked Example: Between a Line and a Parabola

Worked Example: Rotating the Region Between \(y=x\) and \(y=x^2\)

Find the volume formed by rotating the region bounded by \(y=x\), \(y=x^2\), and their intersection endpoints around the \(y\)-axis.

Describe the region and find its bounds. Intersections satisfy \(x=x^2\), so \(x(x-1)=0\) and \(x=0\) or \(x=1\). Between these values, \(x\) is above \(x^2\), since \(x-x^2=x(1-x)\geq 0\). Thus a vertical slice extends from \(y=x^2\) to \(y=x\).

Choose the shell measurements. The vertical slices are parallel to the \(y\)-axis, so they produce shells about that axis. At position \(x\), the radius is \(x\), the height is \(x-x^2\), and the thickness is \(dx\). The bounds are \(0\leq x\leq 1\).

$$ V=2\pi\int_0^1 x(x-x^2)\,dx =2\pi\int_0^1 (x^2-x^3)\,dx. $$

Evaluate and check. An antiderivative of \(x^2-x^3\) is \(\frac{x^3}{3}-\frac{x^4}{4}\), because its derivative is \(x^2-x^3\). Therefore

$$ V =2\pi\left[\frac{x^3}{3}-\frac{x^4}{4}\right]_0^1 =2\pi\left(\frac{1}{3}-\frac{1}{4}\right) =2\pi\left(\frac{1}{12}\right) =\frac{\pi}{6}. $$

The evaluated volume is positive, as expected. The slice height \(x-x^2\) is nonnegative throughout the interval, and differentiating the displayed antiderivative recovers the integrand.

Worked Example: Horizontal Shells About the \(x\)-Axis

Worked Example: A Region Between Two Curves Written as \(x\) in Terms of \(y\)

Find the volume obtained by rotating the region bounded by \(x=y^2\) and \(x=2y\) around the \(x\)-axis.

Describe the region and find its bounds. The curves meet when \(y^2=2y\), so \(y(y-2)=0\), giving \(y=0\) and \(y=2\). For \(0\leq y\leq 2\), \(2y-y^2=y(2-y)\geq 0\). Thus the left boundary is \(x=y^2\) and the right boundary is \(x=2y\).

Choose the shell measurements. Horizontal slices are parallel to the \(x\)-axis, so they produce shells around the \(x\)-axis. At height \(y\), the radius is the distance \(y\) to that axis. The shell height is the horizontal length \(2y-y^2\), and the thickness is \(dy\). The bounds are \(0\leq y\leq 2\).

$$ V=2\pi\int_0^2 y(2y-y^2)\,dy =2\pi\int_0^2 (2y^2-y^3)\,dy. $$

Evaluate and check. An antiderivative is \(\frac{2}{3}y^3-\frac{1}{4}y^4\), since its derivative is \(2y^2-y^3\). Hence

$$ V =2\pi\left[\frac{2}{3}y^3-\frac{1}{4}y^4\right]_0^2 =2\pi\left(\frac{16}{3}-4\right) =2\pi\left(\frac{4}{3}\right) =\frac{8\pi}{3}. $$

The height is nonnegative over the full interval, and the antiderivative differentiates to the expanded integrand. The result is therefore consistent with both the setup and the evaluation.

Worked Example: A Shifted Vertical Axis

Worked Example: Rotating a Line-Parabola Region Around \(x=-1\)

Find the volume formed by rotating the region bounded by \(y=2x\) and \(y=x^2\) around the vertical line \(x=-1\).

Describe the region and find its bounds. The intersections satisfy \(2x=x^2\), so \(x(x-2)=0\), giving \(x=0\) and \(x=2\). On this interval, \(2x-x^2=x(2-x)\geq 0\); thus \(y=2x\) is the upper boundary and \(y=x^2\) is the lower boundary. A vertical slice has height \(2x-x^2\).

Choose the shell measurements. Vertical slices are parallel to the axis \(x=-1\). The radius is the horizontal distance from \(x\) to \(-1\), namely \(x+1\), not simply \(x\). The thickness is \(dx\), and the bounds are \(0\leq x\leq 2\).

$$ V=2\pi\int_0^2 (x+1)(2x-x^2)\,dx. $$

Evaluate and check. Expanding the integrand gives

$$ (x+1)(2x-x^2) =2x^2-x^3+2x-x^2 =-x^3+x^2+2x. $$

An antiderivative is \(-\frac{x^4}{4}+\frac{x^3}{3}+x^2\), whose derivative is \(-x^3+x^2+2x\). Therefore

$$ \begin{aligned} V &=2\pi\left[-\frac{x^4}{4}+\frac{x^3}{3}+x^2\right]_0^2\\ &=2\pi\left(-4+\frac{8}{3}+4\right)\\ &=2\pi\left(\frac{8}{3}\right) =\frac{16\pi}{3}. \end{aligned} $$

The shell radii range from \(1\) to \(3\), so they are positive, and the heights are nonnegative on the stated interval. Differentiation confirms the antiderivative, and the endpoint evaluation gives a positive volume.

A Common Radius Error

In the shifted-axis example, using \(x\) as the radius would measure distance from the \(y\)-axis, not from the actual axis \(x=-1\). That incorrect choice produces

$$ 2\pi\int_0^2 x(2x-x^2)\,dx =2\pi\left[\frac{2}{3}x^3-\frac{1}{4}x^4\right]_0^2 =2\pi\left(\frac{16}{3}-4\right) =\frac{8\pi}{3}. $$

This calculation is internally consistent but uses the wrong geometry: its radius is \(x\) instead of \(x+1\). To correct it, subtract the axis coordinate from the slice coordinate and take the distance. Here that distance is \(\lvert x-(-1)\rvert=x+1\) on \(0\leq x\leq 2\), giving the correct volume \(\frac{16\pi}{3}\). The radius is always a distance, not automatically the coordinate itself.

A Final Setup Check

Before evaluating a shell integral, verify that the slice orientation, radius, height, and bounds all describe the same region. The radius and height should be nonnegative on the integration interval. If a region has multiple components or its boundary changes, a single height formula may not work throughout; use the appropriate piecewise setup established in earlier material. Also make sure each part of the solid is counted once.

1
Locate the axis.
Decide whether vertical or horizontal slices are parallel to it.
2
Find the interval.
Use the region's endpoints or intersections in the slice coordinate.
3
Measure each slice.
Use distance to the axis for the radius and length inside the region for the height.
4
Build and check the integral.
Use \(2\pi\int rh\,d\), confirm nonnegative measurements, and verify the antiderivative and endpoint evaluation.

The shell method is useful because a region that is awkward to describe with cross-sectional disks can sometimes be described by a simple radius and height for each slice. The central skill is not memorizing a formula in isolation; it is translating a sketch into those two measurements and the correct bounds.

Check Your Understanding

For each question, identify the geometry before attempting any integration.

  1. For the region between \(y=3x\) and \(y=x^2\), what equations determine the intersection bounds?
  2. When rotating a region around the \(x\)-axis, which slice orientation produces cylindrical shells, and what is the thickness differential?
  3. A vertical slice at \(x\) is rotated around \(x=4\), with \(x<4\). What is its shell radius?
  4. For horizontal shells, how is the shell height found when the region's left and right boundaries are given as functions of \(y\)?
  5. What geometric error occurs if the radius is taken to be the coordinate \(x\) when the axis is \(x=-1\)?

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