Why Boundary Curves Do Not Automatically Give the Radii
A washer slice is determined by the distances from the axis of rotation to the two edges of a slice through the region. Those distances are the inner and outer radii. A curve’s height or horizontal position gives a radius only after its distance from the axis has been identified.
This distinction matters particularly when the axis is shifted or when the region lies below or to the left of the axis. The upper curve is not always farther from the axis than the lower curve, and the right boundary is not always farther away than the left boundary.
In the examples below, each slice lies entirely on one side of its axis of rotation. When the region and axis are on the same side throughout an interval, the radii can be found by comparing the boundary distances. The ordering should be checked before the washer formula is used.
Radii Are Distances to the Axis
Suppose vertical slices are used, so a slice at \(x\) runs between a lower curve \(y=L(x)\) and an upper curve \(y=U(x)\), with \(L(x)\le U(x)\). If the horizontal axis of rotation is \(y=k\), the distances from the two boundaries to the axis are \(|L(x)-k|\) and \(|U(x)-k|\). Which distance is larger determines which boundary gives the outer radius.
For horizontal slices, suppose a slice at \(y\) runs from a left boundary \(x=A(y)\) to a right boundary \(x=B(y)\), with \(A(y)\le B(y)\). If the vertical axis of rotation is \(x=h\), the distances are \(|A(y)-h|\) and \(|B(y)-h|\). Again, the larger distance gives the outer radius.
How to Read Radii from Upper and Lower Curves
For a vertical slice rotated around \(y=k\), the curves \(U(x)\) and \(L(x)\) describe the slice’s top and bottom edges. If the whole slice lies above the axis, the upper curve is farther from the axis. If the whole slice lies below the axis, the lower curve is farther from the axis. The radii in each case follow from subtracting the axis height from the boundary heights in the appropriate order.
The first row applies when both boundaries are on or above the axis. The second applies when both are on or below it. These formulas are not interchangeable: below the axis, the lower curve is farther away.
Worked Example: Curves Above the X-Axis
Worked Example: Curves Above the X-Axis
Consider the region between \(y=x^2\) and \(y=x+1\) for \(0\le x\le1\). Rotate it about the \(x\)-axis. Find the radii and set up the volume integral.
On this interval, the curves are ordered because \(x+1-x^2=(1-x)(x+1)\ge0\). Both curves are on or above the \(x\)-axis: \(x^2\ge0\) and \(x+1\ge1\). Thus the line is farther from the axis, and the parabola is nearer.
At the sample point \(x=\tfrac12\), the upper boundary is \(\tfrac12+1=\tfrac32\) and the lower boundary is \((\tfrac12)^2=\tfrac14\). Their distances from the axis are \(\tfrac32\) and \(\tfrac14\), confirming the radius labels at that point. The inequalities above establish the same ordering throughout the interval.
The washer area is \(\pi(R^2-r^2)\), so the volume is \[ V=\pi\int_0^1\big((x+1)^2-(x^2)^2\big)\,dx. \] Expanding and integrating gives \[ V=\pi\int_0^1(x^2+2x+1-x^4)\,dx =\pi\left[\frac{x^3}{3}+x^2+x-\frac{x^5}{5}\right]_0^1. \] At \(x=1\), the bracket equals \(\frac13+1+1-\frac15=\frac{32}{15}\); at \(x=0\), it equals \(0\). Therefore \(V=\frac{32\pi}{15}\).
As a check on the setup, at \(x=\tfrac12\) the squared-radius difference is \((\tfrac32)^2-(\tfrac14)^2=\tfrac94-\tfrac1{16}=\tfrac{35}{16}>0\). This agrees with an outer radius larger than the inner radius.
How to Read Radii from Left and Right Boundaries
For a horizontal slice rotated around \(x=h\), the slice extends from \(A(y)\) to \(B(y)\). If it lies to the right of the axis, the right boundary is farther away. If it lies to the left, the left boundary is farther away.
As with upper and lower curves, these are distances, so the formulas use nonnegative differences. A shifted axis is handled in the same way: subtract its coordinate from the boundary coordinates, then compare the distances.
Worked Example: Horizontal Slices to the Right of the Y-Axis
Worked Example: Horizontal Slices to the Right of the Y-Axis
For \(0\le y\le2\), let a region run horizontally from \(x=y\) to \(x=y+2\). Rotate it about the \(y\)-axis. Determine the radii and compute the volume.
Both boundaries are on or to the right of the axis \(x=0\), since \(y\ge0\) and \(y+2\ge2\). At \(y=1\), the left boundary is \(1\) and the right boundary is \(3\). Their distances from the axis are \(1\) and \(3\), so the right boundary gives the outer radius and the left gives the inner radius.
The difference between the boundary coordinates is \((y+2)-y=2>0\) throughout the interval. Since both are nonnegative, their distances preserve that ordering everywhere on the interval. Hence \(R(y)=y+2\) and \(r(y)=y\).
Using horizontal slices, the volume is \[ V=\pi\int_0^2\big((y+2)^2-y^2\big)\,dy =\pi\int_0^2(4y+4)\,dy. \] An antiderivative is \(2y^2+4y\), so \[ V=\pi[2y^2+4y]_0^2 =\pi\big((2\cdot4+4\cdot2)-0\big) =16\pi. \] The radii also check at the endpoints: at \(y=0\), they are \(2\) and \(0\); at \(y=2\), they are \(4\) and \(2\). In both cases the outer radius is at least the inner radius, as required.
Shifted Axes and Boundary Ordering
When the axis is shifted, the boundary coordinates are not the radii themselves. For example, if a horizontal axis is \(y=k\), a boundary at height \(y=c\) is a distance \(|c-k|\) from it. The same distance rule applies to a vertical axis \(x=h\): a boundary at \(x=c\) is \(|c-h|\) away.
It is useful to write out the distances before deciding which is outer. This avoids relying on the visual order of the curves, which may not match their distance order when the region is on the opposite side of the axis.
Worked Example: Upper and Lower Curves with a Shifted Axis
Consider the region between \(y=x^2+1\) and \(y=x+3\) for \(0\le x\le1\), rotated about \(y=-1\). Find the radii and compute the volume.
The upper curve is indeed \(y=x+3\), since \((x+3)-(x^2+1)=x+2-x^2=(2-x)(x+1)>0\) on \([0,1]\). Both curves lie above \(y=-1\). Their distances from that axis are obtained by adding \(1\) to their heights.
At \(x=\tfrac12\), the upper height is \(\tfrac72\), so its distance from \(y=-1\) is \(\tfrac72-(-1)=\tfrac92\). The lower height is \(\tfrac54\), giving distance \(\tfrac54-(-1)=\tfrac94\). Thus the upper curve gives the outer radius at this sample point. The inequality between the curves holds on the full interval, and both are above the axis, so that ordering holds throughout.
The volume is \[ V=\pi\int_0^1\big((x+4)^2-(x^2+2)^2\big)\,dx. \] Expanding each square gives \((x+4)^2=x^2+8x+16\) and \((x^2+2)^2=x^4+4x^2+4\). Therefore \[ V=\pi\int_0^1(-x^4-3x^2+8x+12)\,dx =\pi\left[-\frac{x^5}{5}-x^3+4x^2+12x\right]_0^1. \] The value at \(1\) is \(-\frac15-1+4+12=\frac{74}{5}\); the value at \(0\) is \(0\). Hence \(V=\frac{74\pi}{5}\). At \(x=\tfrac12\), the squared-radius difference is \((\tfrac92)^2-(\tfrac94)^2=\tfrac{81}{4}-\tfrac{81}{16}=\tfrac{243}{16}>0\), consistent with the ordering.
Why the Washer Formula Uses These Radii
At a fixed slice, rotation sweeps the farther edge around the axis to form a disk of radius \(R\). The nearer edge sweeps out a disk of radius \(r\) inside it. The cross-sectional area is the area of the larger disk minus the area of the smaller disk.
When slices are perpendicular to the axis of rotation and the radii depend on \(x\), the volume is \(V=\pi\int_a^b(R(x)^2-r(x)^2)\,dx\). When the slices depend on \(y\), it is \(V=\pi\int_c^d(R(y)^2-r(y)^2)\,dy\). In either case, these formulas require the radii to describe the same slice and the outer radius to be at least the inner radius throughout the interval of integration.
The radius definitions explain why the formula subtracts in this order: the outer disk contains the inner disk, so its area is greater. If a proposed setup produces \(R<r\), the radius labels or the interval’s ordering need to be checked before integrating.
A Common Mistake: Assuming the Upper Curve Is Always Outer
Suppose a vertical slice runs from \(y=-4\) to \(y=-2\) and is rotated about the \(x\)-axis. A tempting but incorrect choice is to call the upper boundary \(y=-2\) the outer edge and use radius \(2\), with radius \(4\) for the lower boundary. That would give the cross-sectional area \[ \pi(2^2-4^2)=-12\pi, \] which is negative and cannot be the area of a washer.
The correction is to measure both distances from \(y=0\). The lower boundary is \(4\) units away, and the upper boundary is \(2\) units away. Therefore \(R=4\) and \(r=2\), giving \[ A=\pi(4^2-2^2)=\pi(16-4)=12\pi. \] The lower curve is outer here because the entire slice is below the axis. The vertical position “upper” does not mean “farther from the axis.”
Checking the Ordering Before Integrating
A sample point is a useful diagnostic: substitute its coordinate into both boundary functions, calculate their distances to the axis, and compare. But one sample point alone does not prove that the same boundary is farther away throughout an interval. Before integrating, also verify the ordering over the whole interval, using the boundary inequalities and the fact that the region stays on the stated side of the axis.
Use upper and lower curves for vertical slices, or left and right boundaries for horizontal slices.
Subtract the axis coordinate from each boundary coordinate and use absolute values when identifying distances.
At a sample point, check which boundary is farther. Then justify that the ordering holds throughout the integration interval.
Call the larger distance \(R\) and the smaller distance \(r\). Confirm \(R\ge r\ge0\) before using the washer formula.
If the region or boundary ordering does not satisfy the assumptions used for a single pair of radius formulas on the entire interval, do not proceed as though those formulas apply everywhere. Establish the appropriate interval and radius ordering first.
Check Your Understanding
For each question, identify the axis and boundary distances before deciding which is the outer radius.
- A vertical slice runs from \(y=1\) to \(y=5\) and rotates about \(y=0\). What are its outer and inner radii?
- A vertical slice runs from \(y=-6\) to \(y=-2\) and rotates about \(y=0\). Which boundary gives the outer radius, and what are the radii?
- A horizontal slice runs from \(x=3\) to \(x=7\) and rotates about \(x=1\). Find its outer and inner radii.
- For curves \(y=L(x)\) and \(y=U(x)\) above the axis \(y=k\), with \(L(x)\le U(x)\), write the outer and inner radius functions.
- Why does checking the radii at one sample point not, by itself, establish their ordering on an entire interval?
