Why an Offset Axis Changes the Radii
When a region is revolved around a horizontal axis, a vertical slice sweeps out a circular cross section. If the axis is a boundary of the region, the cross section may be a disk. If the axis is outside the region, the slice does not reach the axis: it sweeps out a washer, with an outer circular edge and an inner circular hole.
The key measurement is distance from the axis of rotation. A curve’s height above the \(x\)-axis is not automatically its radius. If the axis is \(y=c\), the radius from a point at height \(y\) is \(|y-c|\). Choosing the correct distances—and deciding which is the inner radius and which is the outer radius—is the central step in setting up a washer integral.
Region and Axis Conditions
Suppose a region lies between the graphs \(y=L(x)\) and \(y=U(x)\) for \(a\le x\le b\), where \(L(x)\le U(x)\). Thus, at each \(x\), the vertical slice extends from height \(L(x)\) to height \(U(x)\). We revolve this region around the horizontal line \(y=c\).
For the formulas below, assume the axis lies strictly on one side of the region throughout the interval. If the axis is below the region, then \(c<L(x)\le U(x)\) for every \(x\). If the axis is above, then \(L(x)\le U(x)<c\) for every \(x\). These conditions ensure that the vertical slice is separated from the axis and its rotation has a hole.
| Position of the axis | Outer radius \(R(x)\) | Inner radius \(r(x)\) |
|---|---|---|
| Below the region: \(c<L(x)\) | \(U(x)-c\) | \(L(x)-c\) |
| Above the region: \(U(x)<c\) | \(c-L(x)\) | \(c-U(x)\) |
In both cases, the outer radius reaches the boundary farther from the axis, and the inner radius reaches the boundary nearer to it. Radii are nonnegative lengths, so these formulas use distances rather than signed differences.
Deriving the Washer Formula
Consider one vertical slice at a fixed \(x\). When revolved, its outer edge traces a circle of radius \(R(x)\), while its inner edge traces a circle of radius \(r(x)\). The cross-sectional area is the area of the larger disk minus the area of the hole. Since a disk of radius \(s\) has area \(\pi s^2\), the washer area is \(\pi R(x)^2-\pi r(x)^2\).
The slice has thickness \(dx\). By the slice principle for volume, adding the cross-sectional areas across the interval gives the volume integral. In particular, when the radii vary continuously, the integral sums the washer volumes in the limit as the slices become arbitrarily thin.
The subtraction matters: the solid includes the material between the two circular boundaries, not the entire outer disk. The same formula applies whether the axis lies above or below the region; what changes is the distance used for each radius.
Worked Examples
Worked Example: Region Above an Axis Below It
Revolve the region \(0\le y\le x\), \(0\le x\le2\), around \(y=-1\). At each \(x\), the slice runs from \(y=0\) to \(y=x\). The axis is below the region, so the distance to the upper endpoint is the outer radius and the distance to the lower endpoint is the inner radius:
Substitute these radii into the washer formula:
Check the antiderivative: \(\frac{d}{dx}\left(\frac{x^3}{3}+x^2\right)=x^2+2x\), the integrand. Its endpoint values are \(F(2)=\frac{8}{3}+4=\frac{20}{3}\) and \(F(0)=0\), giving the stated volume.
Worked Example: Region Below an Axis Above It
Revolve the region \(x^2\le y\le4\), \(-2\le x\le2\), around \(y=5\). The axis is above the region. The lower curve \(y=x^2\) is farther from the axis, so it determines the outer radius. The upper curve \(y=4\) determines the inner radius:
The volume is
Evaluate the endpoint values separately:
Therefore,
Check by differentiating the antiderivative: \(F'(x)=24-10x^2+x^4\), which is the expanded washer integrand. The radii also have the expected ordering: \(5-x^2\ge1\) on \([-2,2]\), so \(R(x)\ge r(x)\) throughout.
Worked Example: A Triangular Region Revolved Around an Offset Axis
Revolve the region \(0\le y\le2x\), \(0\le x\le1\), around \(y=3\). The entire region lies below the axis. The lower boundary is farther from \(y=3\), so it gives the outer radius; the upper boundary gives the inner radius:
Hence,
Check: \(\frac{d}{dx}\left(6x^2-\frac{4}{3}x^3\right)=12x-4x^2\). The antiderivative is correct, and its values at \(1\) and \(0\) differ by \(6-\frac43=\frac{14}{3}\). Also, \(r(x)=3-2x\) decreases from \(3\) to \(1\), never becoming negative, while \(R(x)=3\) remains the farther distance.
A Reliable Setup Process
Radius errors are easiest to prevent by separating the geometry from the integration. First identify the slice endpoints and the axis. Then calculate distances from the axis, label the farther distance \(R\) and the nearer distance \(r\), and only then form the integral.
For a vertical-slice model, record its \(x\)-coordinate and the lower and upper heights \(L(x)\) and \(U(x)\).
Check whether \(y=c\) is below or above the entire region on the interval.
Subtract heights from the axis in the appropriate order, or use absolute distances, and identify the farther and nearer endpoints.
Use \(V=\pi\int_a^b(R(x)^2-r(x)^2)\,dx\), with limits covering the full region.
If a region has different upper or lower boundary curves on different parts of its interval, the radii must reflect the boundaries that actually define each slice. The integral should be split at any point where those boundaries change. Likewise, the washer setup above assumes the axis remains outside the region; if it passes through the region, the geometry of the cross section must be reconsidered rather than applying these radius assignments unchanged.
Common Mistake: Treating Heights as Radii
In the first example, a tempting but incorrect calculation is to use the boundary heights \(x\) and \(0\) directly as the radii. That would give
This computes the volume for rotation around \(y=0\), not around \(y=-1\). The axis has shifted every radius by one unit: the correct outer and inner radii are \(x+1\) and \(1\), respectively. The corrected calculation is \(\pi\int_0^2((x+1)^2-1)\,dx=\frac{20\pi}{3}\). A quick diagnostic is to ask whether each proposed radius measures a distance from the stated axis. If it measures a distance from the \(x\)-axis instead, it is not the correct radius for an offset-axis problem.
Why the Radius Choice Matters
The volume depends on squared distances from the axis. A small error in a radius therefore changes the area of every washer, and the resulting error accumulates across the whole interval. In particular, using only the region’s vertical height \(U(x)-L(x)\) does not give a washer’s area: it gives neither the outer radius nor the inner radius.
For an offset axis, keep the geometric meaning in view: the outer radius reaches the boundary farthest from the axis, and the inner radius reaches the boundary nearest to it. Once these distances are correct, the area difference and the volume integral follow directly from the slice principle.
Check Your Understanding
For each question, identify the axis position and reason from distances rather than boundary heights alone.
- A region has vertical slices from \(y=1\) to \(y=4\), and is revolved around \(y=0\). Which endpoint determines the outer radius, and what are the two radii?
- A region has vertical slices from \(y=x^2\) to \(y=3\) on \(0\le x\le1\), and is revolved around \(y=5\). Write \(R(x)\), \(r(x)\), and the volume integral.
- Why is the cross-sectional area \(\pi(R^2-r^2)\), rather than \(\pi(R-r)^2\)?
- For the region \(0\le y\le x\), \(0\le x\le2\), what incorrect axis does the calculation \(\pi\int_0^2x^2\,dx\) describe?
- What geometric condition on the axis and the region is needed for the washer assignments in this tutorial?
