Why One Slice Can Have Several Shell Segments
In a shell setup, a slice of the region is revolved around an axis. If the slice consists of one continuous segment, its height is the distance between its two endpoints. But a slice can also pass through a hole or meet disconnected parts of the region. Then it contains several solid segments separated by empty space.
The shell height must count the solid segments, not the empty gaps between them. In particular, subtracting the lowest boundary from the highest boundary can give the wrong height when the slice is disconnected.
We will describe vertical slices and rotation about the \(y\)-axis. At a fixed \(x\), the slice is parallel to the axis of rotation. If its solid parts occupy several disjoint intervals of \(y\)-values, each interval contributes its own length to the shell height.
Endpoints shared by intervals do not affect their lengths or the volume. If two proposed intervals overlap, first combine them into their union; otherwise their overlap would be counted twice.
Using Total Slice Length in a Shell Integral
For rotation about the \(y\)-axis, a vertical slice at \(x\geq 0\) produces a shell of radius \(x\). The shell volume formula uses the shell’s circumference, height, and thickness. When a slice has multiple solid segments, its height in that formula is the total length of those segments.
If the slice geometry changes with \(x\), define \(h(x)\) separately on the intervals where the boundary segments are consistent, then split the integral at the change points. The change points include places where an interval starts or ends, or where a boundary description changes.
The key geometric task is therefore to describe the slice as a union of disjoint intervals. The shell formula then uses the sum of their lengths.
Why the Heights Add
Consider a narrow vertical strip of width \(\Delta x\) at \(x\). Suppose its slice has several disjoint segments whose lengths add to \(h(x)\). When the strip is revolved, each segment forms a thin shell piece with approximately the same radius \(x\) and thickness \(\Delta x\). Its approximate volume is its circumference times its height times its thickness.
The pieces are separate in the direction parallel to the axis: the gaps between them remain empty. Because the slice intervals are disjoint, their shell pieces do not occupy the same vertical positions. Adding their volumes counts the solid and excludes the gaps. Summing these contributions over narrow strips gives a Riemann sum for the volume integral.
Worked Example: Two Separate Bands with the Same \(x\)-Range
Worked Example: Two Separate Bands with the Same \(x\)-Range
Let the region be the union of the rectangles \(0\leq x\leq 2,\ 0\leq y\leq 1\) and \(0\leq x\leq 2,\ 3\leq y\leq 4\). Rotate it around the \(y\)-axis. Find the volume using shells.
Step 1: Describe a vertical slice. For every \(x\) between \(0\) and \(2\), the slice has two intervals: $$ [0,1]\quad\text{and}\quad[3,4]. $$ Each interval has length \(1\), so $$ h(x)=(1-0)+(4-3)=1+1=2. $$ The gap from \(y=1\) to \(y=3\) is empty and contributes no height.
Step 2: Set up and evaluate the shell integral. The shell radius is \(x\), so $$ V=\int_0^2 2\pi x(2)\,dx =4\pi\int_0^2 x\,dx =4\pi\left[\frac{x^2}{2}\right]_0^2 =4\pi\left(\frac{4}{2}-0\right) =8\pi. $$
Check: Each rectangle rotates into a cylinder with radius \(2\) and height \(1\). One cylinder has volume \(\pi(2)^2(1)=4\pi\). The two cylinders occupy separate \(y\)-ranges, so their volumes add to \(4\pi+4\pi=8\pi\), matching the shell integral.
Worked Example: A Gap Between Curved Boundary Segments
Worked Example: A Gap Between Curved Boundary Segments
For \(0\leq x\leq 1\), let a region consist of the two bands $$ 0\leq y\leq x \qquad\text{and}\qquad x+1\leq y\leq 2x+3. $$ Rotate the region around the \(y\)-axis. Find its volume.
Step 1: Find each segment’s length. The lower segment has length $$ x-0=x. $$ The upper segment has length $$ (2x+3)-(x+1)=x+2. $$ The total slice height is therefore $$ h(x)=x+(x+2)=2x+2. $$
These intervals do not overlap: the lower interval ends at \(y=x\), while the upper one begins at \(y=x+1\). The gap has length \(1\), so it is not included in \(h(x)\).
Step 2: Integrate. For rotation around the \(y\)-axis, the radius is \(x\): $$ \begin{aligned} V &=\int_0^1 2\pi x(2x+2)\,dx\\ &=2\pi\int_0^1(2x^2+2x)\,dx\\ &=2\pi\left[\frac{2x^3}{3}+x^2\right]_0^1\\ &=2\pi\left(\frac{2}{3}+1-0\right)\\ &=\frac{10\pi}{3}. \end{aligned} $$
Check: Compute the contributions of the two bands separately. The lower band gives $$ 2\pi\int_0^1 x(x)\,dx =2\pi\left[\frac{x^3}{3}\right]_0^1 =\frac{2\pi}{3}. $$ The upper band gives $$ 2\pi\int_0^1 x(x+2)\,dx =2\pi\left[\frac{x^3}{3}+x^2\right]_0^1 =2\pi\left(\frac{1}{3}+1\right) =\frac{8\pi}{3}. $$ Their sum is \(\frac{2\pi}{3}+\frac{8\pi}{3}=\frac{10\pi}{3}\), as before.
Worked Example: Slice Geometry Changes Across the Region
Worked Example: Slice Geometry Changes Across the Region
Let the region be the union of the rectangles $$ 0\leq x\leq 2,\quad 0\leq y\leq 1 $$ and $$ 0\leq x\leq 1,\quad 2\leq y\leq 3. $$ Rotate the region around the \(y\)-axis. Find the volume.
Step 1: Write the slice intervals on each \(x\)-range. For \(0\leq x\leq 1\), the slice has two intervals, \([0,1]\) and \([2,3]\). Their total length is $$ h(x)=(1-0)+(3-2)=2. $$ For \(1<x\leq 2\), only the lower rectangle is present, so $$ h(x)=1-0=1. $$ The slice geometry changes at \(x=1\), so the integral must be split there.
Step 2: Set up and evaluate both parts. $$ \begin{aligned} V &=\int_0^1 2\pi x(2)\,dx+\int_1^2 2\pi x(1)\,dx\\ &=4\pi\left[\frac{x^2}{2}\right]_0^1 +2\pi\left[\frac{x^2}{2}\right]_1^2\\ &=4\pi\left(\frac{1}{2}-0\right) +2\pi\left(\frac{4}{2}-\frac{1}{2}\right)\\ &=2\pi+2\pi\left(\frac{3}{2}\right)\\ &=2\pi+3\pi\\ &=5\pi. \end{aligned} $$
Check: The lower rectangle rotates into a cylinder of radius \(2\) and height \(1\), with volume \(4\pi\). The upper rectangle rotates into a cylinder of radius \(1\) and height \(1\), with volume \(\pi\). Their total volume is \(4\pi+\pi=5\pi\), agreeing with the split shell integral.
A Common Mistake: Counting the Outer Span
A tempting but incorrect shortcut is to take the highest \(y\)-value minus the lowest \(y\)-value at each \(x\). That gives the length of the entire span, including any holes. In the first example, the extreme values are \(0\) and \(4\), so this shortcut would give \(h(x)=4\). But the slice contains only two unit-length segments, so its actual height is \(1+1=2\).
Using the incorrect height would give $$ \int_0^2 2\pi x(4)\,dx =8\pi\left[\frac{x^2}{2}\right]_0^2 =16\pi, $$ twice the correct volume \(8\pi\). The extra \(8\pi\) comes from treating the empty gap as solid.
There is a related error when proposed intervals overlap. For example, the intervals \([0,3]\) and \([2,4]\) have lengths \(3\) and \(2\), but adding those lengths gives \(5\), double-counting \([2,3]\). Their union is \([0,4]\), whose length is \(4\). Always describe the slice as disjoint intervals before adding lengths.
A Reliable Setup Procedure
Choose the coordinate held constant and list the boundary intervals that the slice actually occupies.
Write the slice as a union of disjoint intervals, merging any that overlap.
For each interval, subtract its lower endpoint from its upper endpoint, then add those lengths to obtain the shell height.
Use separate formulas on ranges where the number of intervals or their boundaries differ.
Integrate the shell contribution on each range and add the results.
Why This Careful Height Matters
Shell height represents how much of the slice is solid in the direction parallel to the rotation axis. It is not necessarily the distance between the slice’s outermost boundaries. Correctly accounting for multiple segments is essential when the region has holes, separate components, or boundary descriptions that change across the interval.
Before integrating, test the height at a representative value of \(x\). Sketch or list the slice intervals, compute their lengths, and check that no gap has been included and no overlap has been counted twice. That check often reveals a setup error before it affects the integral.
Check Your Understanding
For each question, identify the solid slice intervals before writing a shell height.
- A vertical slice consists of \([0,2]\) and \([5,7]\). What is its total shell height, and how long is the gap?
- For \(0\leq x\leq 2\), a region has slice intervals \([0,x]\) and \([x+2,5]\). Find the total height \(h(x)\).
- A region has two slice intervals for \(0\leq x\leq 1\), but only one for \(1<x\leq 3\). Why should the shell integral be split at \(x=1\)?
- What is wrong with adding the lengths of \([0,3]\) and \([2,4]\) as if they were disjoint?
- For rotation around the \(y\)-axis, write the shell-volume integral for the region in question 2, using \(0\leq x\leq 2\).
