Why Boundary Switches Matter
A shell-volume integral depends on the height of each shell. For vertical shells, height is the top boundary’s \(y\)-value minus the bottom boundary’s \(y\)-value. For horizontal shells, height is the right boundary’s \(x\)-value minus the left boundary’s \(x\)-value.
When curves cross or exchange order, the same named curve may not remain the upper, lower, left, or right boundary across the whole region. A single subtraction used throughout can therefore produce negative heights or describe the wrong region. The remedy is to divide the interval into pieces on which the ordering is fixed.
Find the Ordering Before Writing the Integral
Suppose a region has vertical slices described by several candidate lower boundaries and upper boundaries. At a particular \(x\), the actual lower boundary is the greatest of the lower-boundary values, while the actual upper boundary is the least of the upper-boundary values. Thus, the vertical slice runs from the active lower boundary to the active upper boundary.
For horizontal slices, reverse the comparison: the actual left boundary is the smallest candidate \(x\)-value, and the actual right boundary is the greatest candidate \(x\)-value. These comparisons are local: they must be made on the interval where the region is being considered.
A switch can occur where two candidate boundaries have equal values. To find the relevant switch, solve that equality and retain solutions in the stated domain. Then check the ordering on either side. An equality is only a switch for the region’s shell height if the curves involved actually serve as boundaries there.
Vertical slices lead to vertical shells; horizontal slices lead to horizontal shells.
Determine which curve is upper and lower, or left and right, on each part of the domain.
Use a separate height formula on each interval where the ordering is fixed.
Multiply circumference by height, integrate each piece, and add the results.
Shell Formulas with Changing Boundaries
For a vertical shell at position \(x\), rotated about the \(y\)-axis, the radius is \(x\) when the region lies to the right of the axis. Its circumference is \(2\pi x\), and its height is the upper \(y\)-value minus the lower \(y\)-value. If the ordering changes at \(x=c\), use separate height formulas on the intervals before and after \(c\).
Here \(U_1,L_1\) and \(U_2,L_2\) are the actual upper and lower boundaries on their respective intervals. The same idea applies to horizontal shells about the \(x\)-axis: at height \(y\), the radius is \(y\) when the region is above the axis, and shell height is the right \(x\)-value minus the left \(x\)-value.
These formulas assume the stated shell orientation and that the region’s slices and radii are represented correctly on the integration intervals. If the axis or the region’s position makes the radius different from \(x\) or \(y\), use the appropriate distance to the axis instead.
Worked Example: Vertical Shells with a Top-Bottom Switch
Worked Example: Vertical Shells with a Top-Bottom Switch
The region lies between \(y=x\) and \(y=2-x\) for \(0\le x\le2\). Rotate it about the \(y\)-axis. Find the volume using vertical shells.
First find where the curves have equal heights:
For \(0\le x\le1\), \(2-x\ge x\), so the upper curve is \(y=2-x\) and the lower curve is \(y=x\). The shell height is \((2-x)-x=2-2x\). For \(1\le x\le2\), \(x\ge2-x\), so the upper curve is \(y=x\) and the lower curve is \(y=2-x\). The height is \(x-(2-x)=2x-2\).
The shell radius is \(x\). Therefore:
Check: The height is zero at \(x=1\), where the curves meet, and nonnegative on both intervals. Differentiating the antiderivatives gives \(2x-2x^2\) and \(2x^2-2x\), the corresponding integrands before the factor \(2\pi\). The volume is \(4\pi\) cubic units.
Why the Split Is Necessary
The shell method adds the volumes of thin shells. On a short interval of width \(\Delta x\), a vertical shell has approximate circumference \(2\pi r\), height \(h\), and thickness \(\Delta x\). Its approximate volume is \(2\pi r h\,\Delta x\). Adding these contributions and taking the limit gives the integral of \(2\pi r h\).
That reasoning requires \(h\) to be the actual distance between the region’s boundaries. If a pair of curves switches order, subtracting them in the same order on both sides makes the difference change sign. A negative difference is not a negative shell height; it signals that the subtraction order is wrong on that interval.
For continuous boundary curves, their relative order cannot change on an interval without their values becoming equal somewhere in between. Therefore, once the relevant equalities have divided the domain into intervals, checking the order at a point in each interval identifies the correct boundary ordering throughout that interval, provided no other switch occurs there.
Worked Example: Horizontal Shells with a Left-Right Switch
Worked Example: Horizontal Shells with a Left-Right Switch
The region lies between \(x=y\) and \(x=2-y\) for \(0\le y\le2\). Rotate it about the \(x\)-axis. Find the volume using horizontal shells.
Set the two \(x\)-values equal to locate their switch:
For \(0\le y\le1\), \(y\le2-y\), so the left boundary is \(x=y\) and the right boundary is \(x=2-y\). The shell height is \((2-y)-y=2-2y\). For \(1\le y\le2\), \(2-y\le y\), so the left boundary is \(x=2-y\) and the right boundary is \(x=y\). The height is \(y-(2-y)=2y-2\).
The shell radius is \(y\). Hence:
Check: The left and right boundaries agree at \(y=1\), giving zero shell height there. The heights are nonnegative on their stated intervals. The derivatives of the displayed antiderivatives are \(2y-2y^2\) and \(2y^2-2y\), as required. The volume is \(4\pi\) cubic units.
Worked Example: Both the Upper and Lower Boundary Switch
Worked Example: Both the Upper and Lower Boundary Switch
For \(0\le x\le2\), let the region have lower boundary \(y=\max(0,x-1)\) and upper boundary \(y=\min(2,3-x)\). Rotate the region about the \(y\)-axis. Find its volume using vertical shells.
The lower candidates \(0\) and \(x-1\) are equal at \(x=1\). For \(x<1\), \(0>x-1\), so the greater candidate, and thus the lower boundary of the region, is \(y=0\). For \(x>1\), \(x-1>0\), so the lower boundary is \(y=x-1\).
The upper candidates \(2\) and \(3-x\) are also equal at \(x=1\). For \(x<1\), \(3-x>2\), so the smaller candidate, and thus the upper boundary, is \(y=2\). For \(x>1\), \(3-x<2\), so the upper boundary is \(y=3-x\). The region’s height is therefore \(2\) on \(0\le x\le1\), and \((3-x)-(x-1)=4-2x\) on \(1\le x\le2\).
Using radius \(x\), the volume is:
Check: At \(x=1\), the first height is \(2\), and the second formula gives \(4-2(1)=2\), so the height matches across the switch. At \(x=2\), the second height is zero, as the region closes. Differentiating \(2x^2-\frac{2}{3}x^3\) gives \(4x-2x^2=x(4-2x)\), confirming the second integrand. The volume is \(\frac{14\pi}{3}\) cubic units.
Common Mistake: Keeping the Same Subtraction Order
In the first example, a tempting but incorrect setup is to use \(y=2-x\) as the upper curve and \(y=x\) as the lower curve throughout \(0\le x\le2\). That produces:
A negative volume exposes the error: beyond \(x=1\), \(2-x\) is below \(x\), so the purported height is negative. Taking an absolute value of the final answer would not repair the setup, because it would still combine signed contributions from two different shell heights. The correction is to switch the subtraction order at \(x=1\), as in the first worked example.
A Reliable Check Before Integrating
Before evaluating a shell integral with changing boundaries, inspect every piece. Confirm that the height formula is nonnegative on its interval, that formulas agree at a switch when the same boundaries meet there, and that the radius is the distance from the shell to the axis of rotation. A quick test at one point on each interval often catches a reversed subtraction.
Keep the distinction between the shell’s radius and its height clear: the radius measures distance to the axis, while the height measures the span of the region in the slice direction. Only the height’s boundary ordering changes when the region’s upper/lower or left/right boundaries switch.
Check Your Understanding
For each question, identify the correct boundary ordering before writing or evaluating an integral.
- For vertical shells, curves \(y=3-x\) and \(y=1+x\) cross at \(x=1\). Which curve is upper on \(0\le x\le1\), and which is upper on \(1\le x\le2\)?
- A horizontal slice has candidate boundaries \(x=y\) and \(x=4-y\) on \(0\le y\le4\). At what value of \(y\) do left and right switch?
- If the proposed shell height is negative at an interior point of an interval, what should you check before changing the sign of the integral?
- For the region \(0\le x\le2\) between \(y=0\) and \(y=2-x\), what is the vertical-shell height about the \(y\)-axis?
- In a shell integral, what geometric quantity does the radius represent, and what quantity does the height represent?
