Why Intersections Matter
When a region between two curves is revolved around a horizontal axis, each vertical slice can produce a washer. Its area is the area of the outer disk minus the area of the inner disk. The curves determine those radii, but their roles are not necessarily the same across the whole interval.
At an intersection, the curves have the same height. If they exchange which one is farther from the axis, the outer and inner radii exchange roles there too. A single washer formula with fixed curve assignments can then give an incorrect result, including a negative “area.” The reliable method is to locate the switching points, split the interval, and compare the distances to the axis on each piece.
Radii and the Washer Area
Let \(f\) and \(g\) be continuous boundary functions on \([a,b]\), and revolve the region between their graphs around the horizontal line \(y=c\). In this tutorial, assume both curves lie in the same closed half-plane bounded by that axis: either \(f(x),g(x)\ge c\) for every \(x\), or \(f(x),g(x)\le c\) for every \(x\). Thus each vertical segment between the curves stays on one side of the axis, and its rotation forms a washer (possibly a disk when the inner radius is zero).
The radii are distances, so they are nonnegative. For curves that both stay above the axis, the higher curve is farther from it; for curves that both stay below, the lower curve is farther from it. Describing the radii as distances works in either case and prevents sign errors.
For fixed \(x\), subtracting the two squared distances in the appropriate order gives a useful expression for the area:
The absolute value makes the area nonnegative. When the curves stay on the same side of the axis, their distances are ordered in the same way as their heights if they are above the axis, and in the reverse way if they are below. In either case, the curves can exchange which distance is larger only where they meet.
Partitioning at Finitely Many Intersections
Suppose \(f\) and \(g\) have finitely many interior intersections in \([a,b]\), at \(x_1<x_2<\cdots<x_n\). Set \(x_0=a\) and \(x_{n+1}=b\). Between two consecutive partition points the curves do not meet, so their order cannot change there: if it did, continuity would force an intersection in between. On each piece, identify the farther curve and use its distance as \(R\); use the other distance as \(r\).
Values at the partition points do not affect the integral, so adjacent pieces can share an endpoint. The finite-intersection condition matters: continuity alone does not guarantee that there are finitely many intersections or that a finite partition of this form exists.
Why the Partition Works
The justification has two parts. First, the washer area at each \(x\) is determined by the two distances from the curves to the axis. Second, between consecutive intersections, those distances keep the same order. The integral on each piece therefore adds the cross-sectional areas for that portion of the solid.
Solve \(f(x)=g(x)\) in the interval, and keep the interior solutions as partition points.
Choose a test value inside the piece. The curve farther from \(y=c\) gives \(R\); the other gives \(r\).
Evaluate \(\pi(R^2-r^2)\) on its own interval, then add the resulting volumes.
For the order claim in Step 2, consider the difference of the two distances. It is continuous on each piece. If its sign changed, the Intermediate Value Theorem would give a point where the distances were equal. Since both curves lie in the same half-plane, equal distances imply equal heights, hence an intersection. There are no such points in the open piece, so the order is constant there. This proves that the same curve supplies the outer radius throughout that piece.
Worked Example: Two Curves Exchange Roles Once
Worked Example: Two Curves Exchange Roles Once
Find the volume formed by rotating the region between \(f(x)=x^2\) and \(g(x)=2x\) on \([0,3]\) around the \(x\)-axis.
Both curves are nonnegative on this interval. Find their intersections:
The interior switching point is \(x=2\); \(x=0\) is the left endpoint. Test \(x=1\): \(g(1)=2>f(1)=1\), so \(g\) is outer on \([0,2]\). Test \(x=3\): \(f(3)=9>g(3)=6\), so \(f\) is outer on \([2,3]\). Thus
Evaluate the first integral:
The derivative of \(\frac{4x^3}{3}-\frac{x^5}{5}\) is \(4x^2-x^4\), as required. For the second integral,
The derivative of \(\frac{x^5}{5}-\frac{4x^3}{3}\) is \(x^4-4x^2\). Adding the two pieces gives
Both piece integrals are positive, consistent with nonnegative washer areas. The answer is the sum of the volumes on \([0,2]\) and \([2,3]\), with the outer curve changing at their intersection.
Worked Example: Several Intersections
Worked Example: Several Intersections
Rotate the region between \(f(x)=x^2\) and \(g(x)=1\) on \([-2,2]\) around the \(x\)-axis. Both curves are nonnegative. Their intersections satisfy
For \(-2<x<-1\) and \(1<x<2\), \(x^2>1\), so \(x^2\) gives the outer radius. For \(-1<x<1\), \(1>x^2\), so the constant curve gives the outer radius. The setup is
The first and third integrals are equal because their integrands are even and their intervals are reflections. Compute one:
The derivative of \(\frac{x^5}{5}-x\) is \(x^4-1\). The middle piece is
The derivative of \(x-\frac{x^5}{5}\) is \(1-x^4\). Therefore
The central interval contributes positive area because the constant curve is outer there; the two side intervals contribute positive area because \(x^2\) is outer. This confirms that the radii were ordered consistently on all three pieces.
Worked Example: A Shifted Axis
Worked Example: A Shifted Axis
Rotate the region between \(f(x)=1+x\) and \(g(x)=3-x\) on \([0,3]\) around \(y=0\). Both curves are nonnegative there. Their intersection is found from
At \(x=0\), \(g(0)=3>f(0)=1\), so \(g\) is outer on \([0,1]\). At \(x=2\), \(f(2)=3>g(2)=1\), so \(f\) is outer on \([1,3]\). The volume is
Expand the integrands:
Thus the first integral is \([8x-4x^2]_0^1=4\), and the second is \([4x^2-8x]_1^3=12-(-4)=16\). The displayed antiderivatives differentiate to \(8-8x\) and \(8x-8\), respectively. Hence \(V=20\pi\). Each piece has positive area, as the outer radius has been selected by distance from the axis.
When There Are Infinitely Many Intersections
The finite partition formula requires finitely many interior intersections. Continuous curves can instead intersect infinitely many times, so it is not always possible to list a finite set of switching points and form the stated finite sum. Under the same-side-of-axis assumptions, the cross-sectional area formula still applies at every \(x\). The integrand is continuous, so the volume can be written directly as
For example, on \([0,1]\), let \(f(x)=x\sin(1/x)\) for \(x>0\), \(f(0)=0\), and \(g(x)=0\). These functions are continuous and intersect infinitely many times. Rotating the region between them around \(y=2\) keeps both curves below the axis. The absolute-value integral is applicable, but the finite-intersection sum is not. This distinction prevents a finite partition from being asserted when none is available.
A Common Mistake: Keeping the Same Outer Curve
In the first example, using \(g(x)=2x\) as the outer curve throughout \([0,3]\) would give the incorrect setup
A negative volume signals the mistake: for \(x>2\), \(x^2\) is farther from the axis than \(2x\), so the supposed outer radius is actually the inner one. The correction is to split at \(x=2\), reverse the radii on \([2,3]\), and obtain \(\frac{317\pi}{15}\), as computed above. Never infer that one curve stays outer merely because it is outer at an endpoint.
Check Your Understanding
For each question, identify the boundary distances before deciding which curve is outer.
- Why can two boundary curves exchange outer and inner roles only at an intersection when they lie in the same half-plane relative to the axis?
- For \(f(x)=x^2\) and \(g(x)=2x\) on \([0,3]\), which curve is outer on each side of the interior intersection?
- Why must the radii in a washer area be nonnegative?
- What condition allows the volume to be written as a finite sum over intersection intervals?
- How does the absolute-value integral handle infinitely many intersections?
