Why a Washer Integral May Need to Be Split
A washer cross section has an outer radius and an inner radius. When the region’s boundary curves change order, the curve farther from the axis—and therefore the outer radius—can change as the slice moves across the region.
The washer formula still applies on each slice, but its radii must describe that slice correctly. If one curve is farther from the axis on the first part of an interval and the other curve is farther away on the second part, a single unsplit expression can subtract the radii in the wrong order. The remedy is to locate the switch and integrate separately on each side.
Distances and the Switch Condition
First consider vertical slices rotated about a horizontal line \(y=c\). Suppose the two boundary curves on a slice have heights \(f(x)\) and \(g(x)\). Their distances from the axis are \(|f(x)-c|\) and \(|g(x)-c|\). If the whole slice lies on one side of the axis, these distances are the candidate radii of a washer.
A possible switch occurs where the distances are equal. For boundaries \(f\) and \(g\) about \(y=c\), solve
If both boundaries stay on the same side of the axis, equal distances mean \(f(x)=g(x)\). Their intersection is then a candidate switch. If the curves merely touch and do not exchange which one is farther away, the equality point is not a genuine switch: the same curve supplies the outer radius on both sides. Check the intervals on either side, rather than assuming every solution requires a split.
For horizontal slices rotated about a vertical line \(x=a\), use the same reasoning with boundary coordinates \(L(y)\) and \(U(y)\). The distances are \(|L(y)-a|\) and \(|U(y)-a|\), and a candidate switch satisfies
When the entire slice stays on one side of the axis, the greater of those distances gives \(R\). The order of the curves from left to right alone does not settle which distance is greater if the axis is shifted.
Why Splitting Gives the Correct Volume
The washer method adds the volumes of thin cross sections. On an interval where the radii are determined by fixed boundary curves, the cross-sectional area is \(\pi(R^2-r^2)\), so integrating that area gives the volume on that interval. If the radius roles exchange at \(x=s\), the correct area expression on one side may differ from the one on the other. Additivity of the definite integral then gives the total volume as the sum of the two integrals.
Here \(R_1,r_1\) are the outer and inner radii on \([a,s]\), and \(R_2,r_2\) are those on \([s,b]\). For several genuine switches, make a partition at each one and add the corresponding integrals. The reason for splitting is not a change in the washer formula; it is a change in which distances belong in that formula.
A Reliable Setup
Use vertical slices for rotation about a horizontal axis, or horizontal slices for rotation about a vertical axis.
Measure each boundary from the axis. Use absolute values when needed; a radius is a nonnegative distance.
Solve where the two distances are equal within the interval of integration.
Check which boundary is farther from the axis on each side of each candidate point. Split only where the radius roles actually exchange.
On each interval, put the larger distance in \(R\), the smaller in \(r\), and add the resulting integrals.
Worked Examples
Worked Example: Curves Exchange Roles About the \(x\)-Axis
The region lies between \(y=x\) and \(y=x^2\) for \(0\le x\le2\). Find the volume when it is rotated about the \(x\)-axis.
Both curves are nonnegative on this interval, so each vertical slice stays on or above the axis. Their distances from the axis are \(x\) and \(x^2\). Set them equal to locate the candidate switch:
At \(x=0\) the interval begins. For \(0<x<1\), \(x>x^2\), so \(R=x\) and \(r=x^2\). For \(1<x\le2\), \(x^2>x\), so \(R=x^2\) and \(r=x\). Thus
Antiderivatives are \(x^3/3-x^5/5\) and \(x^5/5-x^3/3\), respectively. Evaluating the first integral gives
For the second integral,
Therefore \(V=\pi(2/15+58/15)=4\pi\) cubic units. Check the antiderivatives: differentiating \(x^3/3-x^5/5\) gives \(x^2-x^4\), and differentiating \(x^5/5-x^3/3\) gives \(x^4-x^2\). The integrands are nonnegative on their assigned intervals, as washer areas must be.
Worked Example: A Shifted Horizontal Axis
The region lies between \(y=x\) and \(y=2-x\) for \(0\le x\le2\). Find the volume when it is rotated about \(y=-1\).
The curves meet where \(x=2-x\), so \(2x=2\) and \(x=1\). Both curves are above \(y=-1\) throughout the interval. Their distances from the axis are \(x+1\) and \(3-x\). Compare them:
On \(0\le x<1\), \(3-x\) is farther from the axis. On \(1<x\le2\), \(x+1\) is farther. Hence
Expand the first integrand:
The first integral without \(\pi\) is \([8x-4x^2]_0^1=8-4=4\). The second integrand is \(8x-8\), so the second integral without \(\pi\) is
Thus \(V=8\pi\) cubic units. Check: the derivatives of \(8x-4x^2\) and \(4x^2-8x\) are \(8-8x\) and \(8x-8\), the respective expanded integrands. At the switch \(x=1\), both radii equal \(2\), confirming that the washer area formulas meet consistently there.
Worked Example: Horizontal Slices About the \(y\)-Axis
The region lies between \(x=y^2\) and \(x=y\) for \(0\le y\le2\). Find the volume when it is rotated about the \(y\)-axis.
Use horizontal slices because the axis is vertical. Both boundary coordinates are nonnegative, so their distances from the \(y\)-axis are \(y^2\) and \(y\). They are equal when
For \(0<y<1\), \(y>y^2\), so \(R=y\) and \(r=y^2\). For \(1<y\le2\), \(y^2>y\), so \(R=y^2\) and \(r=y\). The volume is
Evaluating gives
Therefore \(V=4\pi\) cubic units. Check: differentiating the displayed antiderivatives gives \(y^2-y^4\) and \(y^4-y^2\), respectively. The radii are equal at \(y=1\), and the chosen outer radius is greater than the inner radius on each open interval.
Common Pitfall: Keeping the Same Outer Curve
A frequent error is to label one boundary “outer” for the entire interval just because it is outer at the starting point. In the first example, using \(x\) as the outer radius all the way to \(x=2\) would give
For \(1<x\le2\), however, \(x^2>x\). This expression subtracts the larger squared radius from the smaller one, producing a negative washer area on that part. For instance, at \(x=2\), it uses \(2^2-2^4=4-16=-12\), which cannot be a cross-sectional area. The correction is to split at \(x=1\) and reverse the radii on the second interval.
A related error is to solve \(f(x)=g(x)\) but never check whether the curves exchange roles. Equality gives candidate points; testing intervals establishes the correct assignments. Finally, if a slice crosses the rotation axis, the cross section may be a disk rather than a washer. In that situation the inner-radius subtraction is not appropriate; the axis-crossing case requires the disk setup.
When Splitting Matters
A switch changes the formula for the washer area, not the underlying geometry or the washer method itself. Finding every genuine switch before integrating makes the setup reliable, especially when the axis is shifted or the slice direction changes. It also provides a quick check: on every interval, the chosen outer radius must be at least as large as the inner radius, and the resulting area must be nonnegative.
Check Your Understanding
For each question, identify the relevant distances and explain how you decide whether an integral needs to be split.
- For rotation about \(y=0\), the boundaries are \(y=x\) and \(y=x^2\) on \(0\le x\le3\). At what interior point do the radius roles switch?
- For rotation about \(y=-2\), what distances from the axis correspond to boundaries \(y=f(x)\) and \(y=g(x)\), assuming both curves lie above the axis?
- If two boundary distances are equal at a point, what interval test tells you whether that point is a genuine switch?
- For horizontal slices rotated about \(x=1\), what equation identifies candidate points where two boundary distances are equal?
- Why is an expression with \(R<r\) on part of an interval evidence that the radius assignments or the interval partition need correction?
