Why Horizontal Slices Make Washers
When a planar region is rotated around the y-axis, a horizontal slice is perpendicular to the axis of rotation. As it sweeps around the y-axis, the slice traces a circular ring: a washer. If the region touches the axis, the inner radius is zero and the washer is a disk.
The radii are distances measured horizontally, from the y-axis to the boundaries of the region. In particular, if a horizontal slice runs from \(x=a(y)\) to \(x=b(y)\), with \(0\leq a(y)\leq b(y)\), then its outer radius is \(b(y)\) and its inner radius is \(a(y)\). The variable of integration is \(y\), because the slices are stacked vertically.
This setup assumes the region lies on or to the right of the y-axis and that its horizontal slices have the stated inner and outer boundaries throughout the interval. The formula subtracts the area of the hole from the area of the full circular cross section.
Why the Washer Integral Gives Volume
At a fixed height \(y\), rotating a horizontal segment from \(x=r(y)\) to \(x=R(y)\) around the y-axis sweeps out a circular ring. The outer circle has area \(\pi R(y)^2\); the inner circle, which is removed, has area \(\pi r(y)^2\). Thus the cross-sectional area at height \(y\) is
To obtain the volume, divide the region's vertical range into thin intervals. Over a sufficiently thin interval of height \(\Delta y\), the solid is approximated by a washer of cross-sectional area \(A(y)\), so its volume is approximately \(A(y)\Delta y\). Adding these approximations gives a sum of the form \(\sum A(y_i)\Delta y\). As the interval widths approach zero, this sum converges to the definite integral of the cross-sectional area.
This argument explains both the squared radii and the subtraction: each horizontal cross section is an annulus, not a solid disk.
A Reliable Setup Process
For a region bounded by curves and rotated around the y-axis, first describe the region using horizontal slices. At each height, note the left and right x-coordinates of the slice. Since the axis is \(x=0\), those coordinates give the distances to the two circular boundaries, provided the region is to the right of the axis.
Find the lowest and highest \(y\)-values of the region.
Write its left and right endpoints as functions of \(y\).
The endpoint farther from the y-axis gives the outer radius; the nearer endpoint gives the inner radius.
Use \(\pi\int (R(y)^2-r(y)^2)\,dy\) over the vertical bounds.
Worked Examples
Worked Example: The Region Between \(x=y\) and \(x=y^2\)
Rotate the region between \(x=y\) and \(x=y^2\), for \(0\leq y\leq 1\), around the y-axis. On this interval, \(y\geq y^2\), so the outer radius is \(R(y)=y\) and the inner radius is \(r(y)=y^2\).
Substitute these radii into the washer formula:
An antiderivative is \(y^3/3-y^5/5\), since its derivative is \(y^2-y^4\). Evaluate at the bounds:
Check: differentiating \(y^3/3-y^5/5\) gives \(y^2-y^4\), the integrand. At \(y=1\), the two radii are both \(1\), so the washer area is zero, consistent with the integrand being zero there.
Worked Example: The Region Between \(x=y\) and \(x=\sqrt{y}\)
Rotate the region between \(x=y\) and \(x=\sqrt{y}\), for \(0\leq y\leq 1\), around the y-axis. Since \(\sqrt{y}\geq y\) on this interval, the outer radius is \(R(y)=\sqrt{y}\) and the inner radius is \(r(y)=y\).
The washer integral is
An antiderivative is \(y^2/2-y^3/3\), because its derivative is \(y-y^2\). Therefore,
Check: differentiating \(y^2/2-y^3/3\) gives \(y-y^2\). At both \(y=0\) and \(y=1\), the radii match, so the cross-sectional area is zero at each endpoint, as the integrand indicates.
Worked Example: A Region Between a Line and a Vertical Boundary
Consider the region bounded by \(y=0\), \(x=1\), and \(x=3-y\), and rotate it around the y-axis. The line \(x=3-y\) meets \(x=1\) when \(3-y=1\), so \(y=2\). The vertical bounds are therefore \(0\leq y\leq 2\). At each height, the slice runs from \(x=1\) to \(x=3-y\), so \(r(y)=1\) and \(R(y)=3-y\).
Set up and simplify the washer integral:
An antiderivative is \(y^3/3-3y^2+8y\), since its derivative is \(y^2-6y+8\). Substituting the upper bound gives
At the lower bound, the antiderivative is \(0\). Thus
Check: differentiating the antiderivative gives \(y^2-6y+8=(3-y)^2-1\), the washer integrand. At \(y=2\), both radii equal \(1\), so the cross section closes to zero area there.
A Common Radius Mistake
A washer's radii are not the distance between the two boundary curves. They are the separate distances from each boundary to the axis of rotation. The gap between the boundaries gives the washer's thickness in a horizontal slice, but it does not determine the area by itself.
Worked Example: Why the Gap Is Not a Radius
Take the region bounded by \(x=y\), \(x=2y\), and \(y=1\), including their intersection at \(y=0\), and rotate it around the y-axis. For \(0\leq y\leq 1\), the outer radius is \(2y\) and the inner radius is \(y\). The correct volume is
Check: the derivative of \(y^3\) is \(3y^2\), and the endpoint values are \(1\) and \(0\). A common incorrect calculation takes the gap \(2y-y=y\) as the radius of a disk, giving \(\pi\int_0^1 y^2\,dy=\pi/3\). That is wrong: the cross section has an inner radius \(y\), not zero. The correct washer area is \(\pi((2y)^2-y^2)\), not the area of a disk with radius equal to the gap.
What to Check Before Evaluating
Before integrating, verify that the slice direction matches the axis: around the y-axis, use horizontal slices and integrate with respect to \(y\). Then check that the radii are measured from \(x=0\), and that the outer radius is at least as large as the inner radius over the stated bounds. Finally, differentiate the proposed antiderivative and substitute both bounds carefully.
Check Your Understanding
For each question, identify the radii or the appropriate setup before evaluating.
- For \(0\leq y\leq 1\), a region has horizontal boundaries \(x=y\) and \(x=3y\). When rotated around the y-axis, which is the outer radius, and which is the inner radius?
- Write the washer integral for the region between \(x=1\) and \(x=2-y\), for \(0\leq y\leq 1\), rotated around the y-axis.
- Why does rotation of a slice from \(x=1\) to \(x=4\) produce area \(\pi(4^2-1^2)\), rather than a disk of radius \(3\)?
- Evaluate \(\pi\int_0^1(4y^2-y^2)\,dy\), and check the antiderivative by differentiation.
- For a washer integral around the y-axis, why is the differential \(dy\) rather than \(dx\)?
