From a Slice to a Washer
A cross-section of a solid of revolution depends on the piece of the original region that is rotated. When a slice reaches the axis of rotation, its rotation fills a disk. When the slice stays away from the axis, its rotation leaves an empty space in the middle. The resulting cross-section is a washer: a circular ring with an outer edge and an inner edge.
The important geometric question is not simply how long the slice is. It is how far each end of the slice is from the axis. Those distances determine the washer’s two radii, and the radii determine its area. Getting them in the correct order is essential: the outer radius measures to the farther boundary, and the inner radius measures to the nearer boundary.
What the Radii Measure
A radius is a distance measured perpendicular to the axis of rotation. It is not necessarily the height of a boundary curve, nor is it necessarily the coordinate value of that boundary. For example, if the axis is the line \(y=c\) and a point on a vertical slice has coordinate \(y=d\), its distance from the axis is \(|d-c|\).
Suppose a vertical slice runs between \(y=a(x)\) and \(y=b(x)\), with the axis \(y=c\) below the entire slice. Then the farther endpoint is \(b(x)\), and the nearer one is \(a(x)\). The radii are therefore \(R(x)=b(x)-c\) and \(r(x)=a(x)-c\), where \(R\) denotes the outer radius and \(r\) denotes the inner radius.
Likewise, if a horizontal slice runs between \(x=a(y)\) and \(x=b(y)\), with the axis \(x=c\) to the left of the entire slice, then \(R(y)=b(y)-c\) and \(r(y)=a(y)-c\). In both cases, the distances are taken in the direction perpendicular to the axis.
The area depends on the squares of the radii, not just their difference. In particular, a washer with radii \(5\) and \(2\) does not have the area of a circle of radius \(5-2\). The inner circle must be left out, so its area is accounted for by the \(r^2\) term.
Why Rotation Produces an Annulus
Consider a straight slice whose points all lie on one side of the axis, at distances ranging from \(r\) to \(R\), with \(0<r<R\). Rotating any one point on the slice around the axis carries it along a circle. A point at distance \(\rho\) from the axis traces a circle of radius \(\rho\). As every distance from \(r\) through \(R\) occurs along the slice, the rotating slice fills all the circles with radii between \(r\) and \(R\), including both boundary circles.
It does not fill circles with radii less than \(r\), because the original slice has no points at those distances. Nor does it fill circles with radii greater than \(R\), because the slice has no points that far from the axis. Thus the cross-section is exactly the region between the two concentric circles: a washer. The area of that annulus is the area associated with its outer radius minus the area associated with its inner radius, giving \(A=\pi(R^2-r^2)\).
The slice becomes the segment that is swept around the axis.
Measure perpendicular distances from the axis, not distances from the origin unless the axis is at the origin.
The larger distance is \(R\); the smaller distance is \(r\).
Use \(A=\pi(R^2-r^2)\). The inner radius contributes a subtraction because the central region is absent.
Worked Examples
Worked Example: A Rectangular Region Rotated About a Horizontal Axis
Rotate the rectangle \(1\leq x\leq 3\), \(2\leq y\leq 5\) about the \(x\)-axis. Find the washer radii and the volume.
A vertical slice is perpendicular to the horizontal axis. At every \(x\) between \(1\) and \(3\), the slice runs from \(y=2\) to \(y=5\). The axis is \(y=0\), so the endpoint distances are \(2-0=2\) and \(5-0=5\). Therefore \(r=2\) and \(R=5\).
The cross-sectional area is constant:
The slice variable ranges from \(x=1\) to \(x=3\), so the volume is the integral of these cross-sectional areas:
To check the result, the rotated rectangle has length \(3-1=2\) along the axis and a constant washer cross-section of area \(21\pi\). Its volume is \(2(21\pi)=42\pi\), matching the integral. The hole remains present because the slice never reaches the axis: the inner radius is \(2\), not \(0\).
Worked Example: A Changing Inner Radius
For \(1\leq x\leq 3\), consider the region between \(y=x\) and \(y=4\). Rotate it about the \(x\)-axis. Set up and evaluate the volume.
Use vertical slices. The axis is \(y=0\), below both boundaries on this interval. The farther boundary is \(y=4\), so \(R(x)=4-0=4\). The nearer boundary is \(y=x\), so \(r(x)=x-0=x\). The slice area is
Integrate this area from the left endpoint \(1\) to the right endpoint \(3\):
For an algebra check, the derivative of \(16x-\frac{x^3}{3}\) is \(16-x^2\), the factor inside the area formula. The endpoint areas also make sense: \(A(1)=15\pi\), while \(A(3)=7\pi\). The inner radius grows from \(1\) to \(3\), so the washer narrows. The volume is positive, consistent with the positive slice areas throughout the interval.
Worked Example: Horizontal Slices About a Vertical Axis
For \(0\leq y\leq 2\), consider the region between \(x=y+2\) and \(x=6\). Rotate it about the \(y\)-axis. Find the volume.
The axis \(x=0\) is to the left of both boundaries. Horizontal slices are perpendicular to this vertical axis. The farther boundary is \(x=6\), giving \(R(y)=6-0=6\). The nearer boundary is \(x=y+2\), giving \(r(y)=y+2\). Thus
Integrate with respect to \(y\), using the stated limits:
Differentiate the antiderivative: \(36y-\frac{(y+2)^3}{3}\) has derivative \(36-(y+2)^2\), as required. At \(y=0\), the radii are \(R=6\), \(r=2\), so the area is \(32\pi\). At \(y=2\), they are \(R=6\), \(r=4\), so the area is \(20\pi\). The area decreases across the interval, in agreement with the increasing inner radius.
Keep the Geometry and the Integral Distinct
The washer formula describes one cross-section at a time. If the slice position is \(x\), its area may vary with \(x\); if the slice position is \(y\), its area may vary with \(y\). The volume calculation then adds these areas across the full interval of slice positions. The variable of integration must match the slice direction: vertical slices are integrated with respect to \(x\), and horizontal slices with respect to \(y\).
For a volume, the radii are lengths, so \(R^2-r^2\) has units of area. The area expression \(A=\pi(R^2-r^2)\) therefore has square units. Integrating with respect to a length adds one more length unit, producing cubic units. This is a useful dimensional check on a setup.
A Common Radius Mistake
A frequent error is to subtract the boundary coordinates first and treat the resulting thickness as the washer’s outer radius. In the second example, the distance between the boundaries is \(4-x\), but that is not either radius. The axis is \(y=0\), so the actual radii are \(R=4\) and \(r=x\).
For instance, at \(x=2\), the wrong expression gives \(4\pi\), while the correct area is \(\pi(16-4)=12\pi\). The mistake treats the slice thickness as the radius of a solid disk, rather than measuring both boundaries from the axis. A reliable correction is to mark the axis, measure each endpoint’s perpendicular distance to it, and only then identify the larger and smaller distances.
Summary
A slice that stays away from the axis sweeps out a washer, with the empty center represented by its inner radius. Measure both radii from the axis, assign the farther distance to \(R\) and the nearer distance to \(r\), and use
When the slice position varies across an interval, integrate this cross-sectional area over that interval to obtain the volume. Before integrating, check that the chosen slices are perpendicular to the axis, the radii are distances from the axis, and \(R\geq r\geq 0\).
Check Your Understanding
Answer these questions using the geometry of the slice and its distance from the axis.
- A slice has endpoint distances \(7\) and \(3\) from the axis. What are its outer and inner radii, and what is its area?
- Why does rotating a segment whose nearest point is a positive distance from the axis leave a central hole?
- A vertical slice runs from \(y=1\) to \(y=6\), and the axis is \(y=-2\). What are the two radii?
- For a washer with \(R=5\) and \(r=4\), explain why its area is not \(\pi(5-4)^2\).
- If a solid is described using horizontal slices perpendicular to a vertical axis, which variable should be used in the volume integral?
