Why Use Horizontal Slices?
When a planar region is rotated around a vertical line, horizontal slices are natural candidates for cross-sections: each slice is perpendicular to the axis of rotation. The resulting cross-section is a disk or washer, and its radii come from horizontal distances in the original region.
This orientation is especially useful when the region’s left and right boundaries can be written simply as functions of the vertical coordinate, while describing the region using vertical slices would require solving more complicated equations. The key is to express the region’s horizontal extent at each height before measuring the radii.
Describing a Region by Horizontal Slices
Suppose a region \(R\) extends from \(y=y_0\) to \(y=y_1\), and at each height \(y\) its horizontal slice runs from a left boundary \(x=x_L(y)\) to a right boundary \(x=x_R(y)\). Thus, for every \(y\) in the interval, the slice consists of the points satisfying \(x_L(y)\le x\le x_R(y)\).
For a solid formed by rotating this region around the vertical line \(x=c\), the distances from the slice endpoints to the axis are \(|x_L(y)-c|\) and \(|x_R(y)-c|\). Whether these distances are washer radii or disk radii depends on the position of the axis relative to the slice.
Washer and Disk Areas
If the axis \(x=c\) lies outside the slice, rotating the slice produces a washer. The outer radius \(R(y)\) is the farther endpoint’s distance from the axis, and the inner radius \(r(y)\) is the nearer endpoint’s distance. The area is the area of the outer circle minus the area of the inner circle.
If the axis lies on the slice, rotation produces a disk with radius equal to the farthest endpoint’s distance from the axis. If the axis passes through the interior of the slice, every distance from zero up to that farthest distance occurs, so the cross-section is still a disk, not a washer.
The slice principle for volume says that a solid’s volume is the integral of its cross-sectional area over the coordinate along the axis. Here the slices are at heights \(y\), so the thickness is \(dy\). Consequently, once the area has been identified at each height, the volume is integrated with respect to \(y\).
Deriving the Washer Formula for a Vertical Axis
Consider a horizontal segment from \(x=a\) to \(x=b\), with \(a\le b\), rotated around \(x=c\). If the axis is to the left of or at the segment, then \(c\le a\). The inner radius is \(a-c\), and the outer radius is \(b-c\). The cross-sectional area is therefore \(\pi[(b-c)^2-(a-c)^2]\).
If the axis is to the right of or at the segment, then \(b\le c\). The inner radius is \(c-b\), and the outer radius is \(c-a\), giving area \(\pi[(c-a)^2-(c-b)^2]\). In either case, the outer radius must be at least the inner radius. Applying the slice principle at each height gives the volume integral.
Identify the lowest and highest \(y\)-coordinates of the region being rotated.
Express the left and right boundaries as \(x_L(y)\) and \(x_R(y)\).
Compare each horizontal slice with \(x=c\); determine whether it makes a washer or a disk.
Use \(dy\) and the full height interval to obtain the volume.
Worked Example: A Curved Region Rotated Around an Offset Axis
Worked Example: A Curved Region Rotated Around an Offset Axis
Let \(R\) be the region bounded by \(y=x^2\), \(y=4\), and \(x=0\). Rotate it around the vertical line \(x=-1\). Find the volume.
At height \(y\), the curve \(y=x^2\) gives \(x=\sqrt y\) on the region’s right side. The left boundary is \(x=0\). The heights run from \(y=0\) to \(y=4\), so each horizontal slice extends from \(0\) to \(\sqrt y\). The axis \(x=-1\) is to the left of the entire slice.
Distances from the axis to the endpoints are \(1\) and \(1+\sqrt y\). Therefore the outer radius is \(1+\sqrt y\), and the inner radius is \(1\). The area is
Integrating over the full height interval gives
To check the integration, differentiate the antiderivative: \(\frac{d}{dy}\left(\frac{y^2}{2}+\frac{4}{3}y^{3/2}\right)=y+2\sqrt y\), as required. The boundary slices also agree with the geometry: at \(y=0\), the slice has collapsed to \(x=0\), so both radii are \(1\) and the area is zero. At \(y=4\), the slice runs from \(x=0\) to \(x=2\), giving radii \(1\) and \(3\), and area \(\pi(9-1)=8\pi\).
Worked Example: A Region to the Left of the Axis
Worked Example: A Region to the Left of the Axis
For \(0\le y\le2\), let a region have horizontal slices from \(x=y\) to \(x=3\). Rotate it around \(x=4\). Find the volume.
The axis is to the right of every slice. The endpoint \(x=3\) is nearer to the axis, at distance \(4-3=1\). The endpoint \(x=y\) is farther away, at distance \(4-y\). Thus \(R(y)=4-y\) and \(r(y)=1\), and
Integrating from \(y=0\) to \(y=2\),
The derivative of \(15y-4y^2+\frac{y^3}{3}\) is \(15-8y+y^2\), confirming the antiderivative. At the bottom of the region, \(y=0\), the radii are \(4\) and \(1\), so the area is \(15\pi\). At the top, \(y=2\), the radii are \(2\) and \(1\), so the area is \(3\pi\). These values match the shrinking washer cross-sections.
Worked Example: A Slice That Crosses the Axis
Worked Example: A Slice That Crosses the Axis
For \(0\le y\le1\), let the horizontal slice extend from \(x=y-1\) to \(x=y+2\). Rotate the region around \(x=0\). Find the volume.
For this height range, \(y-1\le0\) and \(y+2>0\), so the axis lies within each slice. The slice reaches distances \(1-y\) to the left and \(y+2\) to the right. Since \(y+2\ge1-y\) for \(0\le y\le1\), its farthest endpoint is on the right. Rotation fills the disk from radius zero to radius \(y+2\).
Therefore,
Differentiating the antiderivative gives \(y^2+4y+4=(y+2)^2\), confirming the integration. At \(y=0\), the disk radius is \(2\), and its area is \(4\pi\). At \(y=1\), the radius is \(3\), and its area is \(9\pi\), consistent with the slice extending from \(x=0\) to \(x=3\).
A Common Radius Mistake
The outer and inner radii are distances from the axis, not simply the right and left \(x\)-coordinates. Their order depends on where the axis lies. In the first example, the slice at \(y=0\) runs from \(x=0\) to \(x=0\), not from \(x=0\) to \(x=2\).
Before forming a washer area, check the slice endpoints at both ends of the \(y\)-interval. Then verify that the proposed outer radius is at least as large as the inner radius. If the axis crosses the slice, do not subtract an inner circle: the cross-section is a disk.
Choosing the Horizontal Orientation
Horizontal slices are useful when the region can be described over a clear interval of \(y\)-values and its left and right boundaries are manageable functions of \(y\). For instance, rewriting \(y=x^2\) as \(x=\sqrt y\) makes the first example’s horizontal bounds direct. In another region, horizontal bounds might be more complicated than the original vertical description, so this orientation would not simplify the setup.
The boundary functions and radii must describe the entire region over the integration interval. If the left or right boundary changes at some height, a single pair of formulas may no longer describe all slices; the limits or formulas must then be adjusted to match the geometry. In every case, check that the slice being rotated belongs to the stated region and that the axis’s position has been assessed at every height.
Check Your Understanding
Use the horizontal-slice method to reason through these questions.
- A slice runs from \(x=1\) to \(x=5\) and is rotated around \(x=0\). What are the inner and outer radii, and is the cross-section a disk or washer?
- A slice runs from \(x=-2\) to \(x=3\) and is rotated around \(x=0\). Is the cross-section a disk or a washer? What is its radius?
- A region has horizontal bounds \(x=0\) and \(x=\sqrt y\) for \(0\le y\le4\), and is rotated around \(x=-1\). Write the volume integral, without evaluating it.
- Why must the outer radius be at least as large as the inner radius in a washer formula?
- For the region in the third worked example, what are the disk radii at \(y=0\) and \(y=1\)?
