Why Vertical-Axis Washers Use Horizontal Slices
When a region is rotated around a vertical axis, a slice perpendicular to that axis is horizontal. Rotating a horizontal slice produces a circular cross section. If the slice does not reach the axis of rotation, that cross section has a hole: it is a washer.
For rotation around the \(y\)-axis, the radius of a washer is a horizontal distance. A boundary written as \(x=g(y)\) gives the \(x\)-coordinate of an edge of the slice, so its distance from the \(y\)-axis is \(|g(y)|\). The volume integral therefore adds washer areas over \(y\), using \(dy\).
The Washer Formula for a Vertical Axis
Suppose a horizontal slice at height \(y\) rotates around the \(y\)-axis. Let \(R(y)\) be the outer radius and \(r(y)\) the inner radius, where \(R(y)\ge r(y)\ge 0\). The cross-sectional area is the area of the outer circle minus the area of the hole:
If the region extends from \(y=a\) to \(y=b\), and those radii describe the washers throughout that interval, its volume is the integral of the cross-sectional area:
The bounds are \(y\)-values because the slices are horizontal. The radii are not the \(y\)-coordinates of the curves: they are horizontal distances from the axis to the slice’s edges. For a region to the right of the \(y\)-axis, if a slice runs from \(x=L(y)\) to \(x=U(y)\), with \(0\le L(y)\le U(y)\), then \(r(y)=L(y)\) and \(R(y)=U(y)\).
Why the Washer Integral Gives Volume
The justification follows from the cross-sectional-area method. At each height \(y\), rotation produces a washer whose area is \(A(y)\). Over a thin vertical thickness \(\Delta y\), the small piece of volume is approximately \(A(y)\Delta y\). Adding these pieces over the interval gives sums of the form \(\sum A(y_i)\Delta y\). As the thicknesses approach zero, these sums approach the definite integral of the cross-sectional area.
The subtraction in the area formula is essential: the inner circle is empty space, not part of the solid. The squared radii appear because the area of a circle with radius \(s\) is \(\pi s^2\).
Worked Setups About the \(y\)-Axis
Worked Example: A Linear Inner Boundary
The region is bounded by \(x=y\), \(x=2\), \(y=0\), and \(y=2\). Rotate it around the \(y\)-axis. Set up and evaluate the volume.
A horizontal slice at height \(y\) runs from \(x=y\) to \(x=2\). Both boundaries are to the right of the \(y\)-axis for \(0\le y\le2\). Thus the outer radius is \(R(y)=2\), and the inner radius is \(r(y)=y\). The volume is
Check the antiderivative: \(\frac{d}{dy}\left(4y-\frac{y^3}{3}\right)=4-y^2\), the integrand. At \(y=2\), the two boundaries meet, so the washer has inner and outer radii both equal to \(2\), giving area zero as expected. The volume is \(\frac{16\pi}{3}\) cubic units.
Worked Example: A Curved Inner Boundary
The region is bounded by \(x=y^2\), \(x=4\), \(y=0\), and \(y=2\). Rotate it around the \(y\)-axis. Set up and evaluate the volume.
At height \(y\), the slice runs from \(x=y^2\) to \(x=4\). Since \(0\le y^2\le4\) on this interval, the outer radius is \(R(y)=4\), and the inner radius is \(r(y)=y^2\). Therefore,
Check the antiderivative: \(\frac{d}{dy}\left(16y-\frac{y^5}{5}\right)=16-y^4\). At \(y=0\), the inner radius is zero, so the cross section is a disk; at \(y=2\), both radii are \(4\), so its area is zero. These endpoint checks agree with the boundaries. The volume is \(\frac{128\pi}{5}\) cubic units.
Worked Example: Both Boundaries Are Vertical Lines or Linear Functions
The region is bounded by \(x=y+1\), \(x=3\), \(y=0\), and \(y=2\). Rotate it around the \(y\)-axis. Set up and evaluate the volume.
For each \(y\) between \(0\) and \(2\), the slice starts at \(x=y+1\) and ends at \(x=3\). Since \(1\le y+1\le3\), both edges are to the right of the axis. Hence \(R(y)=3\) and \(r(y)=y+1\). The integral is
Check the antiderivative: \(\frac{d}{dy}\left(8y-y^2-\frac{y^3}{3}\right)=8-2y-y^2\). At \(y=0\), the radii are \(3\) and \(1\); at \(y=2\), both radii are \(3\), so the washer area decreases to zero at the top boundary. The computed volume is \(\frac{28\pi}{3}\) cubic units.
A Reliable Setup Process
Before integrating, make sure the bounds, slice direction, and radius measurements all match the geometry. The following sequence keeps those decisions explicit.
For rotation around a vertical axis, use horizontal slices, which are perpendicular to the axis.
Determine the lowest and highest heights covered by the region. These become the limits of integration.
At a typical height \(y\), identify the two \(x\)-coordinates that bound the region’s horizontal slice.
For the \(y\)-axis, each radius is a horizontal distance from \(x=0\), not a distance measured vertically.
Use \(V=\pi\int_a^b(R(y)^2-r(y)^2)\,dy\), with \(dy\) because the slices are horizontal.
A Common Mistake: Subtracting Radii Instead of Areas
In the first example, the radii are \(R(y)=2\) and \(r(y)=y\). A tempting but incorrect setup is to subtract the radii and use \(\pi(R-r)\). For that example, this gives
This is not the washer area: the formula \(\pi s^2\) uses the square of a radius. Subtracting the areas of the two circles instead gives the correct setup:
The cross section provides a useful check. At \(y=0\), the hole has radius zero, so the area must be \(4\pi\). The incorrect expression \(\pi(2-0)=2\pi\) does not give that area. The correct expression, \(\pi(2^2-0^2)=4\pi\), does.
What to Check Before You Finish
A washer setup depends on the geometry of each horizontal slice. Confirm that your radii are distances from the rotation axis, that the outer radius is at least as large as the inner radius on the interval, and that the bounds describe the region’s full vertical extent. The formula assumes those radii correctly describe the cross section throughout the interval; if the slice boundaries or their roles change, the setup may need additional care.
Check Your Understanding
For each question, identify the slice geometry before calculating.
- A region has horizontal slices running from \(x=y+2\) to \(x=5\) for \(0\le y\le3\). When it is rotated around the \(y\)-axis, what are \(R(y)\), \(r(y)\), and the volume integral?
- A region has horizontal slices from \(x=y^2\) to \(x=9\) for \(0\le y\le3\). Set up the washer integral around the \(y\)-axis.
- Why does a rotation around a vertical axis use \(dy\) in the washer integral?
- For outer radius \(R(y)=4\) and inner radius \(r(y)=y\), write the cross-sectional area at height \(y\).
- What is wrong with using \(\pi\int_a^b(R(y)-r(y))\,dy\) as a washer-volume formula?
