Why a Shifted Axis Changes the Radius
When a region rotates around a coordinate axis, a radius can often be read directly from a coordinate: the distance from a point \((x,y)\) to the \(x\)-axis is \(|y|\), and its distance to the \(y\)-axis is \(|x|\). If the rotation axis is shifted, the coordinate alone is no longer the radius. The distance must be measured from the shifted line.
For example, a point with vertical coordinate \(y=5\) is \(5\) units above the \(x\)-axis, but it is only \(2\) units above the horizontal line \(y=3\). For disk problems, this distinction determines the radius of each circular cross section.
Distance to a Shifted Axis
The axis is a line, so the relevant distance is measured perpendicular to that line. A horizontal line has vertical perpendiculars, while a vertical line has horizontal perpendiculars. The sign of a coordinate difference tells which side of the axis a point is on; a radius is a distance, so it is nonnegative.
Suppose the boundary is given as \(y=f(x)\) and the rotation axis is horizontal, \(y=c\). A vertical slice is perpendicular to the axis, so its boundary point has coordinates \((x,f(x))\). Its distance to the axis is \(|f(x)-c|\). If the region lies above the axis, this can be written \(f(x)-c\); if it lies below, it can be written \(c-f(x)\).
For rotation around a vertical axis \(x=c\), use horizontal slices. If the boundary is described as \(x=g(y)\), the distance from \((g(y),y)\) to the axis is \(|g(y)-c|\). The expressions without absolute values are useful when the side of the axis is known. The absolute-value form makes the distance correct regardless of that side.
| Rotation axis | Perpendicular slice | Boundary description | Distance to axis |
|---|---|---|---|
| Horizontal \(y=c\) | Vertical | \(y=f(x)\) | \(|f(x)-c|\) |
| Vertical \(x=c\) | Horizontal | \(x=g(y)\) | \(|g(y)-c|\) |
Why These Distance Formulas Work
The formulas follow directly from the definition of perpendicular distance. To find the distance from a point \((x,y)\) to the horizontal line \(y=c\), compare it with an arbitrary point \((t,c)\) on that line. The distance is
The value is smallest when \(t=x\), because then the horizontal difference is zero. The shortest distance is therefore \(\sqrt{(y-c)^2}=|y-c|\). For the vertical line \(x=c\), compare \((x,y)\) with points \((c,t)\). The smallest distance occurs at \(t=y\), giving \(\sqrt{(x-c)^2}=|x-c|\).
When a slice runs continuously from the axis to one boundary, rotating it sweeps out a filled disk. Every point on the boundary of that disk is the same perpendicular distance from the axis, so the disk’s radius is precisely the distance formula above. If the slice does not reach the axis, that reasoning does not give a disk; a disk radius cannot be assigned just by measuring to the far boundary as though the entire disk were filled.
Worked Examples
Worked Example: A Region Rotated Around \(y=2\)
A region is bounded by \(y=2\) and \(y=2+x^2\), for \(-1\leq x\leq 1\). It is rotated around the horizontal line \(y=2\). Find the disk radius and set up and evaluate the volume integral.
Each vertical slice runs from the axis \(y=2\) to the upper boundary \(y=2+x^2\). The radius is the boundary’s vertical distance from the axis:
This is nonnegative on the full interval. At \(x=0\), \(r(0)=0^2=0\), as the two boundaries meet; at \(x=1\), \(r(1)=1^2=1\), matching the one-unit gap between \(y=2\) and \(y=3\).
Using the disk area \(\pi r^2\), the volume is
Check the integration: \(\frac{d}{dx}\left(\frac{x^5}{5}\right)=x^4\). The radius checks at both ends as well: \(r(-1)=1\) and \(r(1)=1\), so the equal endpoint disk areas are consistent with the symmetric region.
Worked Example: A Region Rotated Around \(x=-3\)
A region is bounded by \(x=-3\) and \(x=y^2-3\), for \(-2\leq y\leq 2\). It is rotated around the vertical line \(x=-3\). Find the radius and evaluate the disk volume.
Horizontal slices are perpendicular to this vertical axis. The boundary \(x=y^2-3\) lies to the right of \(x=-3\), so the radius is the horizontal distance from the axis to the boundary:
At \(y=0\), the boundary is \(x=-3\), so \(r(0)=0^2=0\). At \(y=2\) and \(y=-2\), the boundary is \(x=1\), and the horizontal distance from \(x=-3\) to \(x=1\) is \(4\). The formula agrees: \(r(2)=2^2=4\) and \(r(-2)=(-2)^2=4\).
Here \(y\) is the variable of integration because the slices are horizontal. Thus
Check the antiderivative: \(\frac{d}{dy}\left(\frac{y^5}{5}\right)=y^4\). The positive radius values at \(y=\pm2\) also confirm that the boundary is four units from the axis at both ends.
Worked Example: A Region Below a Shifted Horizontal Axis
A region is bounded by \(y=-1\) and \(y=x-1\), for \(0\leq x\leq 2\). It is rotated around \(y=-1\). Find the radius and the volume.
The line \(y=-1\) is the axis and the lower boundary. The other boundary is \(y=x-1\), which lies on or above the axis on this interval. The radius is
At \(x=0\), both boundaries have height \(-1\), so \(r(0)=0\). At \(x=2\), the upper boundary has height \(1\); the distance from \(y=-1\) to \(y=1\) is \(2\), and the formula gives \(r(2)=2\).
Squaring the radius for disk area and integrating over the given \(x\)-interval gives
Check the antiderivative: \(\frac{d}{dx}\left(\frac{x^3}{3}\right)=x^2\). The endpoint radii, \(0\) and \(2\), agree with the distances between the boundaries and the axis.
A Reliable Setup Process
Use the same short sequence of decisions whenever the axis is shifted. The direction of the axis determines the slice direction and the coordinate difference used for the radius.
Write its equation as \(y=c\) for a horizontal axis or \(x=c\) for a vertical axis.
Use vertical slices for a horizontal axis and horizontal slices for a vertical axis.
Describe it as \(y=f(x)\) or \(x=g(y)\), as appropriate.
Use \(|f(x)-c|\) or \(|g(y)-c|\) for distance, then simplify if the boundary is known to lie on one side.
Only then does the slice rotate into a filled disk with that radius.
Common Mistake: Using the Coordinate Instead of the Distance
A frequent error is to use the boundary’s coordinate as the radius, even though the axis is not at coordinate zero. For the boundary \(y=2+x^2\) rotated around \(y=2\), the incorrect calculation would be
That expression measures the boundary’s height above the \(x\)-axis, not its distance from \(y=2\). At \(x=0\), it predicts a radius of \(2\), even though the boundary point is \((0,2)\), which lies on the rotation axis. The correct calculation subtracts the axis height:
For a vertical axis the same correction applies horizontally. If the axis is \(x=-3\), the distance from a boundary at \(x=1\) is \(1-(-3)=4\), not \(1\). Always subtract the axis coordinate from the boundary coordinate, then interpret the result as a nonnegative distance.
Check the Radius Before Integrating
A useful check is to evaluate the radius at a point where the boundary meets the axis. The radius should be zero there. At other points, compare the formula with the visible perpendicular gap between axis and boundary. If the formula gives a negative value, either the subtraction order should be reversed or an absolute value is needed.
Remember that a shifted axis changes the reference point for measuring distance, not the direction of the measurement. Horizontal axes still require vertical distances; vertical axes still require horizontal distances. Keeping those two decisions separate makes disk setups more reliable.
Check Your Understanding
For each question, identify the perpendicular distance before using it as a disk radius.
- A boundary is \(y=4+x^2\), and the axis is \(y=4\). What is the radius in terms of \(x\)?
- A boundary is \(x=y^2-5\), and the axis is \(x=-5\). What is the radius in terms of \(y\)?
- A boundary is \(y=3-x\), and the axis is \(y=1\), on an interval where \(3-x\geq1\). Write a nonnegative radius formula.
- Why are vertical slices perpendicular to a horizontal rotation axis, while horizontal slices are perpendicular to a vertical rotation axis?
- A slice from the axis to a boundary has radius \(r(x)=x+1\) on \(-1\leq x\leq1\). What is its radius at \(x=0\), and what geometric feature does \(r(-1)=0\) indicate?
