Why a Shifted Axis Changes the Radius
A volume of revolution is formed by rotating a two-dimensional region around a line. If that line is not one of the coordinate axes, the radius of each circular slice is not simply the value of a function such as \(y=f(x)\) or \(x=g(y)\). A radius is a distance from the axis of rotation.
For example, the vertical distance from a point with height \(y\) to the horizontal line \(y=c\) is \(|y-c|\). The horizontal distance from a point with coordinate \(x\) to the vertical line \(x=c\) is \(|x-c|\). Identifying those distances correctly is the key step in setting up a disk or washer integral.
The method in this tutorial uses slices perpendicular to the axis of rotation. A vertical axis calls for horizontal slices; a horizontal axis calls for vertical slices. Each slice rotates into a disk or a washer, and integrating its cross-sectional area gives the volume.
Disks, Washers, and Their Radii
A disk is a circular cross section with no hole. Its area is \(\pi R^2\), where \(R\) is its radius. A washer is a circular cross section with a hole. If its outer radius is \(R\) and inner radius is \(r\), its area is the area of the outer circle minus the area of the hole.
The outer radius is not automatically the function called “upper” or “right” in an area problem. It is the boundary farther from the axis. Likewise, the inner radius is the boundary nearer to the axis. If the region is entirely on one side of the axis, these distances are found by subtracting the axis coordinate from the boundary coordinate in the appropriate order.
Choosing the Slices and Integral
To form disks or washers, each slice must be perpendicular to the axis of rotation. Thus, for a horizontal axis \(y=c\), use vertical slices and integrate with respect to \(x\). For a vertical axis \(x=c\), use horizontal slices and integrate with respect to \(y\).
| Axis of rotation | Perpendicular slice | Variable of integration |
|---|---|---|
| Horizontal line \(y=c\) | Vertical | \(x\) |
| Vertical line \(x=c\) | Horizontal | \(y\) |
Suppose a region is between two curves and each perpendicular slice lies entirely on one side of the axis. For a horizontal axis, the radii are vertical distances from \(y=c\) to the slice’s boundaries. For a vertical axis, they are horizontal distances from \(x=c\) to those boundaries. The volume is the integral of the cross-sectional area over the full interval of slices.
If a slice crosses the axis, its rotation has no hole: the cross section is a disk, with radius equal to the greatest distance from the axis to that slice. If this situation changes across the region, split the integral where the slice geometry changes. The examples below use slices entirely on one side of the axis, so the two boundary distances directly give the outer and inner radii.
Why the Washer Integral Gives the Volume
The formula follows from the geometry of each perpendicular cross section. Consider a thin slice whose rotation produces a washer with outer radius \(R\) and inner radius \(r\). Its area is the area of a circle of radius \(R\), less the area of a circle of radius \(r\).
If the slice has thickness \(\Delta x\), its volume is approximately \(A\Delta x\). Adding the volumes of many thin slices approximates the whole solid. In the limit as the maximum slice thickness approaches zero, this sum becomes a definite integral of the cross-sectional area. For horizontal slices, the thickness is \(\Delta y\) instead.
Choose slices perpendicular to the axis. Rotation of each slice produces a disk or washer.
Subtract the hole’s area from the outer circle’s area: \(A=\pi(R^2-r^2)\).
Integrate \(A\) across the full interval of slices to obtain the volume.
This reasoning explains why the squared radii are subtracted inside the integral. Subtracting the radii first would give the thickness of the washer, not its area.
Worked Examples
Worked Example: A Region Rotated Around a Horizontal Line Below It
The region under \(y=\sqrt{x}\), above \(y=0\), and between \(x=0\) and \(x=4\) rotates around \(y=-1\). Set up and evaluate the volume.
The axis is horizontal, so use vertical slices and integrate with respect to \(x\). The slice runs from \(y=0\) to \(y=\sqrt{x}\). Both boundaries lie above \(y=-1\). The farther boundary is \(y=\sqrt{x}\), so its distance from the axis is the outer radius. The nearer boundary is \(y=0\), giving the inner radius:
The \(x\)-bounds are \(0\) and \(4\). Therefore,
Check the antiderivative: differentiating \(\frac{x^2}{2}+\frac{4}{3}x^{3/2}\) gives \(x+2\sqrt{x}\), the integrand. At \(x=0\), both radii are \(1\), so the cross section has area zero. At \(x=4\), the radii are \(3\) and \(1\), giving area \(\pi(9-1)=8\pi\), consistent with the washer formula. The volume is \(\frac{56\pi}{3}\) cubic units.
Worked Example: Horizontal Slices Around a Vertical Line
The region between \(x=y^2\) and \(x=2\), for \(-\sqrt{2}\le y\le\sqrt{2}\), rotates around \(x=3\). Set up and evaluate the volume.
The axis is vertical, so use horizontal slices and integrate with respect to \(y\). At each \(y\), the slice extends from \(x=y^2\) to \(x=2\). Both boundaries are to the left of \(x=3\). The farther boundary is \(x=y^2\), so the radii are
The bounds come from the intersections \(y^2=2\), so \(y=-\sqrt{2}\) and \(y=\sqrt{2}\). The volume is
For the endpoint calculation, \(y^3=2\sqrt{2}\) and \(y^5=4\sqrt{2}\) when \(y=\sqrt{2}\); the antiderivative is odd, so its value at \(-\sqrt{2}\) is the negative of its value at \(\sqrt{2}\). Differentiating the antiderivative gives \(8-6y^2+y^4\), confirming the integration. At \(y=0\), the radii are \(3\) and \(1\); at either endpoint, both radii are \(1\). These checks agree with the region’s geometry.
Worked Example: Comparing Distances Around a Shifted Horizontal Axis
The region between \(y=x^2\) and \(y=x\), for \(0\le x\le1\), rotates around \(y=2\). Set up and evaluate the volume.
The axis is horizontal, so vertical slices give \(dx\). On this interval, \(x^2\le x\), and both curves lie below \(y=2\). The lower curve \(y=x^2\) is farther from the axis, while \(y=x\) is nearer. Thus,
The volume is
Differentiating the antiderivative gives \(4x-5x^2+x^4\), as required. At \(x=0\), both radii equal \(2\), so the washer has zero area. At \(x=1\), both radii equal \(1\), again giving zero area. The curves meet at both endpoints, so these checks are consistent with the slices closing up there.
A Common Radius Error
In the third example, a common mistake is to call the radius from \(y=x\) the outer radius merely because \(y=x\) is the upper curve. That would use \(R=2-x\) and \(r=2-x^2\). The resulting integrand is negative on the interior of the interval:
A cross-sectional area cannot be negative. The correction is to compare distances from the axis, not to assign radii by “upper” and “lower” labels. Since \(x^2\le x<2\), the distance \(2-x^2\) is greater than \(2-x\). Therefore \(2-x^2\) is the outer radius and \(2-x\) is the inner radius.
A Reliable Setup Routine
For a disk or washer problem around a shifted axis, a careful setup can be organized into a few checks:
- Write down the axis of rotation, including its equation \(y=c\) or \(x=c\).
- Choose slices perpendicular to that line: vertical for a horizontal axis, horizontal for a vertical axis.
- Find the interval of slices from the region’s bounds or the relevant intersections.
- Express both radii as distances from the axis to the slice boundaries.
- Identify the farther boundary as the outer radius and the nearer boundary as the inner radius.
- Use \(\pi(R^2-r^2)\) and check that the area is nonnegative throughout the interval.
The distance step is essential: for a horizontal axis, subtract \(c\) from the boundary’s \(y\)-coordinate and account for which side of the axis the boundary lies on. For a vertical axis, do the same using \(x\)-coordinates. Once the radii and bounds are correct, the integral records the volume one perpendicular slice at a time.
Check Your Understanding
For each question, identify the slice direction and radii before writing the integral.
- The region between \(y=0\) and \(y=x\), for \(0\le x\le2\), rotates around \(y=-1\). What are \(R(x)\) and \(r(x)\)?
- The region between \(x=0\) and \(x=y+2\), for \(-1\le y\le1\), rotates around \(x=4\). Which variable should you integrate with respect to, and what are the radii?
- Why must a slice crossing the axis of rotation produce a disk rather than a washer with a positive inner radius?
- A proposed washer integrand is \(\pi(r^2-R^2)\), where \(R>r\). What is wrong with this ordering?
- For rotation around \(x=c\), what kind of slices make disks or washers, and what is the integration variable?
