When the Axis Passes Through the Region
A shifted axis outside a region gives each parallel strip a distance from the axis on the same side. If the axis passes through the region, strips on opposite sides can be equally far from it. Rotating both strips produces shells at the same radius, so adding their contributions independently may count the same part of the solid more than once.
The key is to organize the region by distance from the axis. At each distance, identify every strip that reaches that radius, then combine the intervals along the axis that those strips sweep out. If the intervals overlap, count their union rather than adding their lengths.
Radius and the Intervals Contributing to a Shell
For rotation about the vertical line \(x=a\), a vertical strip at \(x\) has radius \(r=|x-a|\). Thus, for a given \(r>0\), the possible strip positions are \(x=a-r\) and \(x=a+r\). Either position may lie outside the region, and both may lie inside it.
Each valid vertical strip has an interval of \(y\)-values in the region. Rotating the strip sweeps out a shell extending across that interval along the axis. If both positions are valid, the shell at radius \(r\) occupies the union of their \(y\)-intervals. Denote the total length of that union by \(H(r)\). The shell method then gives the volume by integrating circumference times this total axial length.
Here \(R\) is the greatest distance from the region to \(x=a\); \(H(r)\) is the total length of the union of all valid \(y\)-intervals at \(x=a-r\) and \(x=a+r\). If the region has several separate vertical components, include each valid interval in the union.
For rotation about the horizontal line \(y=c\), use horizontal strips. At radius \(r\), the possible positions are \(y=c-r\) and \(y=c+r\). Their valid \(x\)-intervals are combined in the same way. If \(W(r)\) is the total length of their union, then
In either formula, the integration variable is distance from the axis, so \(r\) is nonnegative. The axial length is a union length, not automatically the sum of lengths from opposite sides.
Why the Union of Intervals Gives the Right Volume
Fix a radius \(r\). Every point of the region at distance \(r\) from the axis sweeps out a circular path of circumference \(2\pi r\). Along the axis, the paths that occur fill one or more intervals. If those intervals overlap, the corresponding circular shell is already present on the overlapping part; including it a second time would not add new volume.
For a small radial thickness \(\Delta r\), a shell of axial length \(L\) has volume approximately \(2\pi rL\,\Delta r\). At fixed \(r\), the appropriate \(L\) is the total length of the union of the axial intervals. Summing these thin contributions and taking the limit gives \(2\pi\int rL(r)\,dr\). This is the shell formula with the overlap handled correctly. The radius-zero boundary contributes no volume.
Worked Examples
Worked Example: A Rectangle Rotated About an Interior Vertical Line
Let \(R\) be the rectangle \(0\le x\le2,\ 0\le y\le3\). Rotate it about \(x=1\), which passes through its interior. Set up and evaluate the shell integral.
At radius \(r\), the possible strip positions are \(x=1-r\) and \(x=1+r\). Both lie in \(0\le x\le2\) when \(0\le r\le1\). At either position, the \(y\)-interval is \([0,3]\). The intervals coincide, so their union has length \(H(r)=3\), not \(6\). No point of the rectangle is more than \(1\) unit from the axis, so \(R=1\).
Check: the rectangle extends \(1\) unit to either side of its axis, and rotating it fills a cylinder of radius \(1\) and length \(3\). Its volume is \(\pi(1)^2(3)=3\pi\), agreeing with the shell integral.
Worked Example: A Rectangle Rotated About an Interior Horizontal Line
Let \(R\) be the rectangle \(0\le x\le4,\ 0\le y\le2\). Rotate it about \(y=1\). Set up and evaluate a shell integral using the strips above the axis.
For \(0\le r\le1\), the strip above the axis is at \(y=1+r\), which lies in the rectangle. Its \(x\)-interval is \([0,4]\), of length \(W(r)=4\). The strip below the axis at \(y=1-r\) also produces radius \(r\), but it sweeps out the same axial interval. Counting only the upper strip therefore accounts for the shell once. The greatest radius is \(1\).
Check: the rotation fills a cylinder whose axis has length \(4\) and whose radius is \(1\). Its volume is \(\pi(1)^2(4)=4\pi\), as computed.
Worked Example: A Triangular Region About a Vertical Line Through Its Interior
Let \(R\) be bounded by \(y=0\), \(y=x\), \(x=0\), and \(x=3\). Rotate it about \(x=1\). Find the shell integral, taking account of how the valid strips change with radius.
At a distance \(0\le r\le1\), both positions \(x=1-r\) and \(x=1+r\) are in the region. The corresponding vertical intervals are \([0,1-r]\) and \([0,1+r]\). The first is contained in the second, so their union is \([0,1+r]\), with length \(H(r)=1+r\).
For \(1<r\le2\), \(x=1-r<0\) is outside the region, while \(x=1+r\) lies between \(2\) and \(3\). Only the latter strip contributes, and its interval is \([0,1+r]\), again giving \(H(r)=1+r\). No point is farther than \(2\) from \(x=1\). Thus the integral must be split at \(r=1\), where the number of valid strip positions changes.
Check: for every \(0<r\le2\), the greatest \(y\)-value among the valid strips is \(1+r\), and all their intervals begin at \(y=0\). Their union therefore has length \(1+r\) throughout. The two pieces of the integral cover the full radial range, with no omitted or duplicated interval. Their evaluated contributions are \(5\pi/3\) and \(23\pi/3\), which sum to \(28\pi/3\).
A Common Double-Counting Error
In the rectangle of the first example, a tempting but incorrect calculation is to add the full contribution from each side of the axis:
This counts the same cylindrical shell twice: at every radius between \(0\) and \(1\), both vertical strips sweep out the same interval \(0\le y\le3\). The rotated region is a cylinder of volume \(3\pi\), not \(6\pi\). The correction is to take the union of the intervals at each radius. Here that union has length \(3\), so the correct setup is \(2\pi\int_0^1 3r\,dr=3\pi\).
Adding lengths is appropriate only when the intervals at that radius are disjoint. For example, if the two valid axial intervals are \([0,1]\) and \([2,4]\), their union has total length \(1+2=3\). If instead they are \([0,2]\) and \([1,4]\), the union is \([0,4]\), of length \(4\), not the sum \(2+3=5\).
A Reliable Setup Checklist
For a vertical axis, use vertical strips; for a horizontal axis, use horizontal strips, parallel to the axis.
Use \(x=a-r,\ x=a+r\) for \(x=a\), or \(y=c-r,\ y=c+r\) for \(y=c\).
Use the region's bounds to determine the radial intervals on which each position contributes.
Combine the valid \(y\)-intervals for a vertical axis or the valid \(x\)-intervals for a horizontal axis. Count overlaps once.
Use circumference \(2\pi r\) times the total union length, and confirm the bounds cover each distance from the axis that occurs in the region.
The shell setup is determined by the solid formed, not by how many original strips generate a particular shell. Tracking both the valid strip positions and the union of their axial intervals makes that distinction explicit and prevents an otherwise plausible integral from counting volume twice.
Check Your Understanding
For each question, identify the radius, the valid strip positions, and the axial interval or intervals that contribute.
- A rectangle \(0\le x\le4,\ 0\le y\le2\) is rotated about \(x=2\). For \(0\le r\le2\), what is the union length \(H(r)\)?
- A region is rotated about \(y=3\). What are the two possible horizontal strip positions at radius \(r\)?
- At one radius, two contributing axial intervals are \([0,3]\) and \([2,5]\). What total length should the shell use?
- Why is adding the shell contributions from both sides of an interior axis incorrect when their axial intervals coincide?
- For the triangular region bounded by \(y=0\), \(y=x\), \(x=0\), and \(x=3\), why must the radial setup be split at \(r=1\) when rotating about \(x=1\)?
