Introduction
A disk-method problem does not require the rotation axis to be the \(x\)-axis or the \(y\)-axis. The same idea works when the axis is a horizontal or vertical line shifted away from the origin. The essential change is geometric: the radius is the distance from the slice’s boundary to the actual rotation axis.
For example, if a vertical slice runs from \(y=c\) to \(y=f(x)\), rotating that segment around \(y=c\) makes a disk of radius \(|f(x)-c|\). Using \(|f(x)|\) instead would measure from the \(x\)-axis, which is a different line.
When Does a Slice Make a Disk?
A disk is the solid circular cross-section with no hole in its middle. To get a disk by rotating a slice, the slice must run from the rotation axis to a boundary of the region. If the slice is a segment from the axis to a curve, rotating it sweeps out a filled circle. The segment’s length is the disk radius.
The slice must be perpendicular to the axis of rotation. Thus, for a horizontal axis, use vertical slices and integrate with respect to \(x\). For a vertical axis, use horizontal slices and integrate with respect to \(y\).
| Rotation axis | Perpendicular slice | Radius from boundary to axis | Volume setup |
|---|---|---|---|
| \(y=c\) | Vertical, thickness \(dx\) | \(|f(x)-c|\) | \(\pi\int_a^b |f(x)-c|^2\,dx\) |
| \(x=c\) | Horizontal, thickness \(dy\) | \(|g(y)-c|\) | \(\pi\int_p^q |g(y)-c|^2\,dy\) |
In the first row, the region must have vertical slices running from \(y=c\) to \(y=f(x)\) over \(a\le x\le b\). In the second, it must have horizontal slices running from \(x=c\) to \(x=g(y)\) over \(p\le y\le q\). The displayed absolute values express distance; if the boundary stays on one side of the axis, the squared radius can also be written as the square of the corresponding signed difference.
These setups assume that each slice reaches the axis and fills the disk. If a slice leaves a gap between the axis and the region, it does not produce a disk.
Why the Shifted-Axis Formula Works
The disk method adds the areas of thin cross-sections along the solid. A disk of radius \(R\) has area \(\pi R^2\). For a horizontal axis \(y=c\), a vertical slice at \(x\) rotates into a disk whose radius is the vertical distance from \(y=c\) to \(y=f(x)\), namely \(|f(x)-c|\). Its area is therefore \(\pi|f(x)-c|^2\).
Over a short width \(\Delta x\), that cross-section contributes approximately \(\pi|f(x)-c|^2\Delta x\) to the volume. Adding these contributions over a partition and taking the limit gives the integral. The vertical-axis argument is the same: a horizontal slice at \(y\) has radius \(|g(y)-c|\), area \(\pi|g(y)-c|^2\), and thickness \(dy\).
These formulas require the described disk-shaped slices over the whole interval. They also require the boundary functions and bounds to describe the region correctly. The integral sums cross-sectional areas, so the variable of integration must match the direction in which the slices are stacked.
Worked Examples
Worked Example: Rotation Around a Horizontal Line Above the Region
Let the region be bounded by \(y=x\), \(y=5\), \(x=0\), and \(x=2\). Find the volume when it is rotated around \(y=5\).
The axis is horizontal, so use vertical slices and integrate with respect to \(x\). Each slice runs from \(y=x\) to \(y=5\), and its rotation about \(y=5\) makes a disk. The radius is the distance between these \(y\)-coordinates:
Therefore the volume is:
Check the antiderivative by differentiating: \(\frac{d}{dx}(25x-5x^2+x^3/3)=25-10x+x^2=(5-x)^2\). The bounds give the upper value \(50-20+8/3=98/3\) and the lower value \(0\), confirming \(V=98\pi/3\) cubic units.
Worked Example: Rotation Around a Shifted Vertical Line
Let the region lie between \(x=1\) and \(x=y^2\) for \(1\le y\le2\). Find the volume when it is rotated around \(x=1\).
The axis is vertical, so use horizontal slices and integrate with respect to \(y\). Each slice runs from the axis \(x=1\) to \(x=y^2\). The radius is the horizontal distance between them:
Thus:
Check by differentiation: \(\frac{d}{dy}(y^5/5-2y^3/3+y)=y^4-2y^2+1=(y^2-1)^2\). Evaluating the antiderivative at \(2\) gives \(46/15\), and at \(1\) gives \(8/15\); their difference is \(38/15\). The volume is \(38\pi/15\) cubic units.
Worked Example: A Curved Boundary Above a Shifted Horizontal Axis
Consider the region between \(y=1\) and \(y=5-x^2\) for \(0\le x\le2\). Find the volume when it is rotated around \(y=1\).
Vertical slices are perpendicular to the horizontal axis. A slice runs from \(y=1\) to \(y=5-x^2\), so its radius is:
The radius is nonnegative on this interval, reaching zero at \(x=2\). Integrating the disk areas gives:
Differentiate the antiderivative: \(16x-8x^3/3+x^5/5\) has derivative \(16-8x^2+x^4=(4-x^2)^2\). Its values at \(2\) and \(0\) are \(256/15\) and \(0\), respectively, confirming \(V=256\pi/15\) cubic units.
A Common Radius Mistake
A frequent error is to use the boundary’s coordinate as the radius, as though the axis were still a coordinate axis. For example, in the first worked example the axis is \(y=5\), and the curve is \(y=x\). The radius is not \(x\); it is the distance from \(y=x\) to \(y=5\).
Before integrating, identify the two coordinates whose difference measures the distance to the axis. For rotation around \(y=c\), compare \(y\)-coordinates. For rotation around \(x=c\), compare \(x\)-coordinates. A radius is a nonnegative length, even when a signed coordinate difference is negative.
Choosing and Checking the Setup
A reliable setup follows the geometry of the slices rather than the appearance of the curve. First identify the axis and choose perpendicular slices. Then verify that each slice runs from that axis to the region’s boundary. Only after those checks should you write the disk area and integrate.
Write it as \(y=c\) for a horizontal line or \(x=c\) for a vertical line.
Use vertical slices and \(dx\) for a horizontal axis; use horizontal slices and \(dy\) for a vertical axis.
Use the absolute difference of the relevant coordinates, such as \(|f(x)-c|\) or \(|g(y)-c|\).
Use \(A=\pi R^2\) and integrate over the slice bounds.
Differentiate the antiderivative and verify the bounds and the units of volume.
The disk method is useful whenever a solid is built from circular slices that reach the axis of rotation. Shifting the axis changes the radius calculation, but not the area formula or the principle of adding cross-sectional areas. Keeping those two ideas separate—how the radius is measured and how disk areas are accumulated—makes shifted-axis problems systematic.
Check Your Understanding
For each question, identify the radius before attempting the integral.
- A region has vertical slices from \(y=2\) to \(y=f(x)\), where \(f(x)\ge2\). What is the disk radius when the region is rotated around \(y=2\)?
- For a region with horizontal slices from \(x=3\) to \(x=g(y)\), where \(g(y)\ge3\), which variable should the volume integral use, and what is the radius?
- A curve is \(y=x+1\), and the rotation axis is \(y=4\). Write the radius of a vertical slice at \(x\), assuming the curve lies below the axis on the interval considered.
- What geometric feature of a slice must hold for its rotation to produce a disk rather than a cross-section with a central gap?
- For rotation about \(x=c\), why is \(|g(y)-c|\) a distance even if \(g(y)-c\) is negative?
