When a Region Meets the Axis of Rotation
A washer is formed when a slice rotates around an axis that lies outside the slice: the rotation leaves a hole. If the slice reaches or crosses the axis, the rotation fills that hole, producing a disk instead. This change matters because subtracting an inner radius in a slice that reaches the axis would incorrectly remove part of the solid.
We will describe the geometry using vertical slices and a horizontal axis \(y=k\). At a fixed \(x\), suppose the region occupies all the \(y\)-values in an interval \([a(x),b(x)]\), where \(a(x)\le b(x)\). The slice rotates around \(y=k\). Its shape is determined by whether \(k\) lies inside that interval.
Distances Determine the Radii
For a horizontal axis \(y=k\), the distance from a point at height \(y\) to the axis is \(|y-k|\). Thus the two slice boundaries are at distances \(|a(x)-k|\) and \(|b(x)-k|\). Whether these distances are inner or outer radii depends on the slice's position relative to the axis.
- If \(a(x)<k<b(x)\), the slice crosses the axis. Rotation produces a disk whose radius is the larger of the two endpoint distances.
- If \(k\) is an endpoint, the slice touches the axis. Rotation still produces a disk, with radius equal to the distance of the other endpoint from the axis.
- If \(b(x)<k\), the slice lies below the axis. The outer radius is \(k-a(x)\), and the inner radius is \(k-b(x)\).
- If \(k<a(x)\), the slice lies above the axis. The outer radius is \(b(x)-k\), and the inner radius is \(a(x)-k\).
For a disk of radius \(R(x)\), the cross-sectional area is \(\pi R(x)^2\). For a washer with outer radius \(R(x)\) and inner radius \(r(x)\), its area is \(\pi(R(x)^2-r(x)^2)\). Integrating these areas over the \(x\)-interval gives the volume.
Why Crossing the Axis Fills the Disk
The disk-versus-washer rule follows directly from the distances covered by a slice. Consider one fixed slice, an interval \([a,b]\), and an axis \(y=k\). As points \(y\) in the slice rotate around the axis, they sweep out circles with radii \(|y-k|\).
If \(a\le k\le b\), the slice contains the axis, so radius zero occurs.
Moving continuously from \(k\) to either endpoint produces every distance from zero up to that endpoint's distance.
The two endpoint distances are \(k-a\) and \(b-k\). Their maximum is the outer radius, so the cross section is a full disk.
In particular, if the axis is inside the slice, there is no missing range of radii between zero and the outer edge. A washer has a hole only when the slice stays away from the axis: then its distances begin at a positive inner radius and extend to the outer radius.
The volume integral may need separate pieces because the slice can lie on one side of the axis for some \(x\)-values and cross it for others. The relevant split occurs where a boundary reaches the axis, such as where \(a(x)=k\) or \(b(x)=k\).
Worked Examples
Worked Example: A Region Changes from Disks to Washers
Rotate the region between \(y=x-1\) and \(y=2\), for \(0\le x\le3\), around the \(x\)-axis. Find the volume.
At each \(x\), the vertical slice runs from \(y=x-1\) to \(y=2\). The axis is \(y=0\). The lower boundary crosses the axis when
For \(0\le x\le1\), the slice contains \(y=0\), so its rotation is a disk. The distances from the axis to the endpoints are \(1-x\) and \(2\); since \(2\ge1-x\) on this interval, the disk radius is \(2\). For \(1\le x\le3\), the slice lies above the axis, so it forms a washer with outer radius \(2\) and inner radius \(x-1\). Therefore
Evaluate the disk contribution:
For the washer contribution, an antiderivative of \(4-(x-1)^2\) is \(4x-\frac{(x-1)^3}{3}\). Thus
Adding the pieces gives \(V=4\pi+\frac{16\pi}{3}=\frac{28\pi}{3}\) cubic units. As a check, the integrand changes at \(x=1\): the disk radius is \(2\), and the washer's inner radius starts at \(0\). Both formulas give cross-sectional area \(4\pi\) there, as they should.
Worked Example: A Slice Crosses the Axis from the Other Side
Rotate the region between \(y=-3\) and \(y=x\), for \(-1\le x\le2\), around the \(x\)-axis. Find the volume.
The vertical slice has endpoint heights \(-3\) and \(x\). For \(-1\le x<0\), both endpoints are below the axis, so the slice makes a washer. Its outer radius is \(3\), and its inner radius is \(|x|=-x\). For \(0\le x\le2\), the slice contains the axis, so it makes a disk of radius \(3\). The split is at \(x=0\), where \(y=x\) meets the axis.
The first integral is
The second integral is
Hence \(V=\frac{26\pi}{3}+18\pi=\frac{80\pi}{3}\) cubic units. At the split \(x=0\), the washer's inner radius becomes zero, so its area agrees with that of a disk of radius \(3\). This confirms the transition is handled consistently.
Worked Example: Crossing a Shifted Axis
Rotate the region between \(y=-1\) and \(y=x\), for \(0\le x\le2\), around the horizontal line \(y=1\). Find the volume.
The axis \(y=1\) meets the upper boundary \(y=x\) at \(x=1\). For \(0\le x<1\), the entire slice lies below the axis, so it produces a washer. The outer radius is the distance from \(y=1\) to \(y=-1\), namely \(2\). The inner radius is the distance from \(y=1\) to \(y=x\), namely \(1-x\). For \(1\le x\le2\), the slice reaches or crosses the axis. The endpoint distances are \(2\) and \(x-1\); since \(x-1\le1<2\) on this range, the disk radius is \(2\). Thus
On the first interval, \(4-(1-x)^2=4-(1-2x+x^2)=3+2x-x^2\). Therefore
The disk contribution is \(\pi\int_1^2 4\,dx=4\pi\). Hence \(V=\frac{11\pi}{3}+4\pi=\frac{23\pi}{3}\) cubic units. At \(x=1\), the washer's inner radius is zero and the disk radius is \(2\), so both expressions give area \(4\pi\).
A Common Mistake: Keeping an Inner Radius When the Slice Crosses the Axis
Suppose a slice extends from \(y=-1\) to \(y=2\) and rotates around the \(x\)-axis. A tempting but incorrect calculation is to call the distances \(1\) and \(2\) the inner and outer radii and write \(\pi(2^2-1^2)=3\pi\). This treats the slice as though it stayed on one side of the axis. It does not: the interval \([-1,2]\) contains \(0\).
The correct cross section is a disk of radius \(2\). The slice includes the axis and points at every distance from \(0\) to \(2\), so no hole remains. Its area is \(\pi(2^2)=4\pi\). The correction is not to subtract the smaller endpoint distance: when a slice crosses the axis, its nearer endpoint does not determine an inner radius.
Using the Same Geometry with a Vertical Axis
The reasoning does not depend on the axis being horizontal. For a vertical axis \(x=h\), use horizontal slices instead. If a horizontal slice covers \(x\)-values in \([c(y),d(y)]\), compare that interval with \(h\). When \(c(y)\le h\le d(y)\), its rotation is a disk, with radius \(\max\{|c(y)-h|,|d(y)-h|\}\). When the interval is entirely to one side, it rotates into a washer. As before, split the integral where a boundary meets the axis and changes the slice geometry.
The essential habit is to inspect the whole slice before assigning radii. First locate the axis relative to the interval; then decide whether the interval rotates into a filled disk or a washer with a genuine hole. The usual disk or washer area formula applies only after that geometric decision.
Check Your Understanding
For each question, decide what the slice geometry implies before calculating.
- A vertical slice runs from \(y=-2\) to \(y=5\) and rotates around the \(x\)-axis. Is its cross section a disk or washer, and what is its radius?
- A slice runs from \(y=1\) to \(y=4\) and rotates around \(y=0\). Give its inner and outer radii.
- A region has slice interval \(y\in[x-2,3]\) and rotates around the \(x\)-axis. At what \(x\)-value does the lower boundary meet the axis?
- Why is it incorrect to subtract the smaller endpoint distance when the slice interval contains the axis?
- A slice runs from \(y=-4\) to \(y=x\) and rotates around \(y=0\). For which \(x\)-values does it contain the axis?
