One Solid, Two Ways to Slice It
A solid of revolution can often be described using either cylindrical shells or washers. The methods slice the same region in different directions, so their integrals generally have different variables, bounds, and integrands. When both setups are valid, they must give the same volume.
The practical question is not which method is universally better. It is which description makes the geometry and calculation clearer. A shell setup may use the original function directly, while a washer setup may require its inverse. In another problem, washers may avoid awkward shell heights or the need to account for symmetry.
The Two Volume Setups
For a thin cylindrical shell with radius \(r\), height \(h\), and thickness \(\Delta x\), its approximate volume is circumference times height times thickness. Adding these contributions and taking a limit gives the shell formula. The differential is taken along the direction in which the shells are stacked.
For washers, each cross-section is a disk or a washer perpendicular to the axis of rotation. If \(R\) is the outer radius and \(r\) is the inner radius, the cross-sectional area is \(\pi(R^2-r^2)\). Integrating those areas along the axis gives the volume.
In either formula, radii are distances from the axis, so they are nonnegative. The shell height is the length of the region along a shell; it is not automatically the value of a function. The bounds must cover the entire solid exactly once.
Worked Example: A Region Under a Power Curve
Worked Example: Rotate \(y=x^2\) About the \(y\)-Axis
Let the region lie under \(y=x^2\), above the \(x\)-axis, and between \(x=0\) and \(x=2\). Rotate it about the \(y\)-axis. Find the volume using both methods.
Shell setup. A vertical strip at \(x\) makes a shell of radius \(x\), height \(x^2\), and thickness \(dx\). The strip runs from \(x=0\) to \(x=2\), so
Evaluating,
Washer setup. A horizontal slice at height \(y\) extends from \(x=\sqrt y\) to \(x=2\). Rotating it about the \(y\)-axis makes a washer with outer radius \(2\) and inner radius \(\sqrt y\). The height runs from \(y=0\) to \(y=4\), giving
Thus,
Check. Both methods give \(8\pi\), so the two integrals agree. Here the shell integrand is a single power, while the washer integrand simplifies because the inner radius is squared. Either setup is short.
Worked Example: Curves That Change Roles in the Two Setups
Worked Example: The Region Between \(y=x\) and \(y=x^2\)
Consider the region between \(y=x\) and \(y=x^2\) for \(0\le x\le1\), rotated about the \(x\)-axis. On this interval, \(x\ge x^2\), so the line forms the upper boundary and the parabola forms the lower boundary.
Washer setup. Vertical slices are perpendicular to the axis. Their outer radius is \(x\), and their inner radius is \(x^2\). The bounds are the original \(x\)-bounds:
Integrating each term,
Shell setup. Horizontal slices are parallel to the axis. At height \(y\), the left boundary is \(x=y\), from the line. The right boundary is \(x=\sqrt y\), from \(y=x^2\). Thus the shell height is \(\sqrt y-y\), and its radius is \(y\). Since \(y\) ranges from \(0\) to \(1\),
Therefore,
Check. The washer and shell calculations both give \(2\pi/15\). The washer setup uses the original curves directly. The shell setup needs the curves rewritten as \(x=y\) and \(x=\sqrt y\), but it still has one interval and a manageable integrand.
Why the Two Integrals Agree
The equality can be seen directly for a common class of regions. Suppose a region is bounded by \(y=f(x)\), the \(x\)-axis, and the vertical lines \(x=a\) and \(x=b\), where \(f\) is continuously differentiable, strictly increasing, \(f(a)=0\), and \(f'(x)>0\) for \(a<x<b\). Assume \(0\le a<b\). These conditions ensure the region lies on the nonnegative side of the \(y\)-axis and allow its boundary to be written as \(x=g(y)\), the inverse of \(f\). Write \(H=f(b)\).
When this region is rotated about the \(y\)-axis, the shell integral is \(2\pi\int_a^b x f(x)\,dx\). A horizontal slice at height \(y\) runs from \(x=g(y)\) to \(x=b\). It therefore produces a washer with outer radius \(b\) and inner radius \(g(y)\), giving \(\pi\int_0^H(b^2-g(y)^2)\,dy\). To verify that these are equal, compare the integrals without their common factor of \(\pi\):
Let \(I=\int_0^H(b^2-g(y)^2)\,dy\). The inverse satisfies \(g(0)=a\) and \(g(H)=b\).
Since \(\frac{d}{dy}(b^2-g(y)^2)=-2g(y)g'(y)\), integration by parts gives \(I=[y(b^2-g(y)^2)]_0^H+\int_0^H2y\,g(y)g'(y)\,dy\). The boundary term is zero: at \(y=H\), \(g(H)=b\), and at \(y=0\), the factor \(y\) is zero.
In the remaining integral, set \(x=g(y)\), so \(y=f(x)\) and \(dx=g'(y)\,dy\). As \(y\) runs from \(0\) to \(H\), \(x\) runs from \(a\) to \(b\). Thus \(I=\int_a^b2x f(x)\,dx\).
Multiplying both expressions by \(\pi\) proves that the shell and washer volumes agree under these hypotheses. If the inverse derivative is unbounded at an endpoint, as for \(f(x)=x^2\) at \(x=0\), the change of variables is understood on the interior and the endpoint is handled by a limit. The improper integral converges in the example above.
Worked Example: Symmetry and a Shared Axis
Worked Example: Rotate \(y=4-x^2\) About the \(y\)-Axis
Take the region above the \(x\)-axis and below \(y=4-x^2\), for \(-2\le x\le2\), and rotate it about the \(y\)-axis.
Washer setup. At height \(y\), the parabola gives \(x^2=4-y\), so the region extends from \(-\sqrt{4-y}\) to \(\sqrt{4-y}\). Rotating this horizontal segment produces a disk, not a washer with a hole. Its radius is \(\sqrt{4-y}\). The height ranges from \(0\) to \(4\):
Shell setup. Each vertical strip at a positive \(x\) has radius \(x\) and height \(4-x^2\). The negative-\(x\) half of the region produces the same shells when rotated. To avoid counting those shells twice, use just the right half, \(0\le x\le2\):
Check and pitfall. Both methods give \(8\pi\). Integrating the shell contributions over the entire interval \([-2,2]\) would count each physical shell twice: using the distance \(|x|\) as radius would give \(2\pi\int_{-2}^2|x|(4-x^2)\,dx=16\pi\). The correction is to integrate over one half of the region, or otherwise account for the overlap, rather than treating the two sides as distinct shells.
How to Choose Between the Methods
Both integrals represent the same solid, but one may be simpler to set up or evaluate. First identify the axis and consider slices in both directions. Then compare the actual bounds and expressions each choice requires.
| What to compare | Shells | Washers |
|---|---|---|
| Slice direction | Parallel to the axis of rotation | Perpendicular to the axis of rotation |
| Typical geometric factors | Radius times shell height | Outer radius squared minus inner radius squared |
| Bounds | Along the direction in which shells are stacked | Along the direction in which cross-sections are stacked |
| Potential complication | Shell heights may require inverse functions or care about overlapping shells | Outer and inner radii may require inverse functions or piecewise bounds |
A useful choice is often the method that keeps the region’s boundaries in their original form and gives a single interval. But that is a guide, not a rule: an inverse function may be simple, while an apparently direct setup may split into several pieces. Symmetry also matters. In the third example, washers naturally describe the disk at each height, while shells require using only one side to avoid duplication.
Check Your Understanding
For each question, use the slice geometry to explain your setup before evaluating any integral.
- For the region under \(y=x^2\) on \(0\le x\le2\), rotated about the \(y\)-axis, what are the shell radius and height? What are the outer and inner washer radii?
- For the region between \(y=x\) and \(y=x^2\), rotated about the \(x\)-axis, why does a horizontal shell at height \(y\) extend from \(x=y\) to \(x=\sqrt y\)?
- In the region under \(y=4-x^2\) above the \(x\)-axis, rotated about the \(y\)-axis, why would integrating shell contributions over both halves count the same shells twice?
- What information should you compare before deciding whether shells or washers give the simpler setup?
- In the general inverse-function derivation, why does the boundary term in the integration-by-parts step vanish?