Why the Axis’s Position Matters
A slice of a region does not always make a washer when rotated. If the slice lies entirely on one side of the axis, its rotation leaves a hole and forms a washer. If the slice reaches or crosses the axis, the rotation fills in to the center and forms a disk. A single region can have both kinds of slices, so its volume integral may need to be split.
The key is to compare the whole slice with the axis before assigning radii. Measuring endpoint distances alone is not enough: when the axis lies between the endpoints, those distances describe the two sides of one filled disk, not the outer and inner radii of a washer.
How to Classify a Slice
Consider a vertical slice whose \(y\)-coordinates run from \(L(x)\) to \(U(x)\), where \(L(x)\le U(x)\). Suppose the region is rotated about the horizontal line \(y=c\). The slice lies below the axis if \(U(x)<c\), above it if \(L(x)>c\), and reaches or crosses it if \(L(x)\le c\le U(x)\). A slice that only touches the axis at an endpoint belongs to the disk case: it has no gap between the rotated slice and the axis.
The same rule applies to horizontal slices rotated about a vertical line \(x=c\): compare the slice’s \(x\)-coordinates with \(c\). In either orientation, the area of a washer with outer radius \(R\) and inner radius \(r\) is \(\pi(R^2-r^2)\), and the area of a disk with radius \(R\) is \(\pi R^2\). The radii are distances, so they are nonnegative.
Why a Crossing Slice Makes a Disk
The classification follows from the distances that occur along the slice. For a vertical slice from \(L(x)\) to \(U(x)\) and axis \(y=c\), the distance from a point \(y\) on the slice to the axis is \(|y-c|\).
As \(y\) moves along the slice, \(|y-c|\) runs from a positive nearest distance \(r\) to a farther distance \(R\). Rotation therefore produces all radii from \(r\) through \(R\), forming an annulus: a washer.
The distance \(|y-c|\) takes the value zero and every value up to the farthest endpoint’s distance. Rotation therefore fills all radii from zero to \(R\), forming a disk.
This also explains why the disk radius is \(\max(c-L(x),\,U(x)-c)\) when \(L(x)\le c\le U(x)\). The two quantities are the distances from the axis to the lower and upper endpoints, and the larger one reaches farthest from the axis.
Worked Example: A Slice Crossing the Axis
Worked Example: A Slice Crossing the Axis
A vertical slice runs from \(y=-1\) to \(y=2\). Rotate it about the \(x\)-axis, \(y=0\). Classify the cross-section and find its area.
The axis value \(0\) lies between \(-1\) and \(2\), so the slice crosses the axis. The endpoint distances are
The farther endpoint is at distance \(2\), so the slice forms a disk with radius \(R=2\). Its area is
Check: the slice contains points at every distance from \(0\) to \(2\). For example, distance \(1\) occurs at \(y=-1\) and \(y=1\), and the axis itself supplies distance \(0\). Rotation therefore fills the disk of radius \(2\); there is no hole.
When to Split the Volume Integral
If the axis crosses only some slices, the cross-sectional area formula changes where the crossing begins or ends. For vertical slices rotated about \(y=c\), find candidate split points by solving \(L(x)=c\) or \(U(x)=c\). These equations locate where an endpoint meets the axis. Then check the intervals between those points to determine whether each slice is a washer or a disk.
For a disk slice, also compare the two endpoint distances. If the farther endpoint changes within an interval, the expression for the radius changes there too, so split at the equality \(c-L(x)=U(x)-c\). The split is needed because the radius is the greater of those two distances; using just one endpoint beyond the point where it stops being farther would give incorrect areas.
Once each interval has one applicable area formula, integrate each formula over its own bounds and add the results. The thickness is \(dx\) for vertical slices and \(dy\) for horizontal slices. This is the cross-sectional volume approach: add the areas of thin slices across the region.
Worked Example: Washer Slices Followed by Disk Slices
Worked Example: Washer Slices Followed by Disk Slices
The region is bounded by \(y=0\), \(y=x\), \(x=0\), and \(x=\frac32\). Rotate it about the horizontal line \(y=1\). Set up and evaluate the volume.
At each \(x\), the vertical slice runs from \(y=0\) to \(y=x\). To find where the upper endpoint meets the axis, solve \(x=1\). Thus the slices for \(0\le x<1\) lie below the axis, and for \(1\le x\le\frac32\) they reach or cross it.
For \(0\le x<1\), the farther endpoint is \(y=0\), at distance \(1\); the nearer endpoint is \(y=x\), at distance \(1-x\). These slices form washers with \(R=1\) and \(r=1-x\). Their area is
For \(1\le x\le\frac32\), the slice crosses the axis. Its endpoint distances are \(1\) and \(x-1\). Since \(x-1\le\frac12<1\) on this interval, the lower endpoint is always farther from the axis. Each slice is a disk of radius \(1\), with area \(A(x)=\pi\).
Integrate the washer areas from \(0\) to \(1\), and the disk areas from \(1\) to \(\frac32\):
Check the integrations by differentiating: \(\frac{d}{dx}\left(x^2-\frac{x^3}{3}\right)=2x-x^2\), and \(\frac{d}{dx}(x)=1\). The split also matches the geometry: before \(x=1\), the slice misses the axis, while at and after \(x=1\), the axis is in the slice.
Worked Example: A Washer When the Axis Does Not Cross
Worked Example: A Washer When the Axis Does Not Cross
A region has \(0\le x\le2\) and \(1\le y\le3\). Rotate it about \(y=0\). Find its volume.
Every vertical slice lies above the axis. The endpoint distances are \(1\) and \(3\), so each slice forms a washer with \(R=3\) and \(r=1\). The cross-sectional area is constant:
The \(x\)-interval has length \(2\), so
Check: the rotated region is an annular cylinder of length \(2\) and cross-sectional area \(8\pi\). Its volume is \(2(8\pi)=16\pi\), agreeing with the integral.
A Common Mistake: Calling a Crossing Slice a Washer
For the slice from \(y=-1\) to \(y=2\) rotated about \(y=0\), a tempting but incorrect calculation is to use endpoint distances \(2\) and \(1\) as washer radii:
That calculation assumes the slice stays at least distance \(1\) from the axis. It does not: the slice contains the axis, so its rotation includes the center and leaves no hole. The correction is to use a disk of radius \(2\):
The reliable habit is to locate the axis relative to the entire slice first. Only after classifying that slice should you assign its disk or washer radii.
Check Your Understanding
For each question, decide how the slice or region rotates before calculating.
- A slice runs from \(y=2\) to \(y=5\) and rotates about \(y=1\). Is the cross-section a disk or a washer, and what are its radii?
- A slice runs from \(y=-3\) to \(y=1\) and rotates about \(y=0\). Is there a hole? Find the disk radius if there is no hole.
- A region’s vertical slices run from \(y=0\) to \(y=2x\) and rotate about \(y=2\). At what \(x\)-value does the upper endpoint meet the axis?
- Why is it not enough to subtract the squares of the two endpoint distances when the slice crosses the axis?
- For a slice from \(y=-2\) to \(y=1\) rotated about \(y=0\), which endpoint determines the disk radius?
