Why Use Horizontal Washers?
When a region is revolved around a vertical line, a horizontal slice is perpendicular to the axis of rotation. Rotating that slice produces a disk or washer. Adding the cross-sectional areas over the region’s vertical extent therefore gives a volume integral with respect to \(y\).
The boundary curves are often supplied as equations of the form \(y=f(x)\). A horizontal slice, however, needs the left and right \(x\)-coordinates of the region at a fixed height \(y\). The key step is to express those boundaries as \(x\)-functions of \(y\), either by solving for \(x\) or by using an equation already written in that form.
Horizontal Boundaries and Inverse Functions
Suppose a boundary is given by \(y=f(x)\). To use it as the endpoint of a horizontal slice, solve the equation for \(x\) in terms of \(y\). If \(f\) is one-to-one on the part of the curve being used, this gives its inverse function \(x=f^{-1}(y)\).
Some curves are not one-to-one across their entire domains. For instance, \(y=x^2\) has two possible solutions for \(x\): \(x=\sqrt{y}\) and \(x=-\sqrt{y}\) when \(y\ge 0\). A particular region may use only one of these branches. Select the branch that actually forms its left or right boundary; do not assume that solving for \(x\) gives just one boundary.
The \(y\)-limits describe the vertical range of the region. They come from the heights where its boundaries meet or from horizontal boundaries that close the region. For each \(y\) in that range, identify the slice’s left and right endpoints before measuring either radius.
Radii Are Horizontal Distances
Let the axis of rotation be the vertical line \(x=c\). A point with horizontal coordinate \(x\) is \(|x-c|\) units from that axis. Thus, the radius is a distance, not simply the \(x\)-coordinate of a boundary.
When the entire slice lies on one side of the axis, its two boundary distances determine the outer and inner radii. The outer radius is the larger distance; the inner radius is the smaller. If the axis lies to the left of the region, the right boundary is farther away. If the axis lies to the right, the left boundary is farther away.
This formula uses the area of an annulus at each height: the area of its outer disk, \(\pi R(y)^2\), minus the area of its inner disk, \(\pi r(y)^2\). Summing these areas over the vertical interval gives the integral in \(y\). The geometry matters: if a proposed inner radius is larger than the outer one, the boundary distances have been assigned incorrectly.
A Setup Process
Determine the lowest and highest \(y\)-values in the region.
At a representative height \(y\), find the left and right \(x\)-coordinates. Solve for \(x\) where needed and select the boundary branches used by the region.
Use horizontal distances from \(x=c\) to the two endpoints, then label the larger distance \(R(y)\) and the smaller one \(r(y)\).
Substitute the radii into \(\pi\int_a^b(R(y)^2-r(y)^2)\,dy\).
Worked Examples
Worked Example: Inverting a Curve to Find the Slice Endpoints
The region between \(y=x^2\) and \(y=x\) for \(0\le x\le 1\) is revolved around the \(y\)-axis. Set up and evaluate the volume using horizontal washers.
First find the range of heights. The curves meet where \(x^2=x\), so \(x(x-1)=0\), giving \(x=0\) and \(x=1\). Their corresponding heights are \(y=0\) and \(y=1\). Thus, \(0\le y\le 1\).
At a fixed height, solve each curve equation for \(x\). From \(y=x\), we get \(x=y\). From \(y=x^2\), the branch in the region \(0\le x\le1\) is \(x=\sqrt{y}\). For \(0\le y\le1\), \(y\le\sqrt{y}\), so the horizontal slice runs from \(x=y\) to \(x=\sqrt{y}\).
The axis is \(x=0\). The inner radius is the distance from \(x=0\) to \(x=y\), and the outer radius is the distance to \(x=\sqrt{y}\): $$ r(y)=y,\qquad R(y)=\sqrt{y}. $$
Therefore, $$ V=\pi\int_0^1\left((\sqrt{y})^2-y^2\right)\,dy =\pi\int_0^1(y-y^2)\,dy. $$ An antiderivative is \(\frac{y^2}{2}-\frac{y^3}{3}\). Evaluating at the bounds, $$ V=\pi\left[\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1 =\pi\left(\frac12-\frac13\right) =\frac{\pi}{6}. $$
Check: the integrand \(y-y^2=y(1-y)\) is nonnegative on \(0\le y\le1\), as required for washer area. Differentiating the antiderivative gives \(y-y^2\), so the evaluation matches the integrand. The volume is \(\frac{\pi}{6}\) cubic units.
Worked Example: A Shifted Axis and Two Inverse Boundaries
The region between \(y=x^2\) and \(y=x\) for \(0\le x\le1\) is now revolved around \(x=-1\). Find the volume with horizontal washers.
As in the preceding setup, the curves meet at \((0,0)\) and \((1,1)\), so the \(y\)-limits are \(0\) and \(1\). Solving for \(x\) gives the horizontal endpoints \(x=y\) and \(x=\sqrt{y}\). Since \(y\le\sqrt{y}\) on this interval, these are the left and right boundaries, respectively.
The axis \(x=-1\) lies to the left of the whole region. The horizontal distance to the right boundary is larger, so $$ R(y)=\sqrt{y}-(-1)=1+\sqrt{y}, \qquad r(y)=y-(-1)=1+y. $$
Substitute these radii into the washer formula: $$ V=\pi\int_0^1\left((1+\sqrt{y})^2-(1+y)^2\right)\,dy. $$ Expand each square: $$ (1+\sqrt{y})^2=1+2\sqrt{y}+y,\qquad (1+y)^2=1+2y+y^2. $$ Thus, $$ V=\pi\int_0^1\left(2\sqrt{y}-y-y^2\right)\,dy. $$ An antiderivative is \(\frac{4}{3}y^{3/2}-\frac{y^2}{2}-\frac{y^3}{3}\). Therefore, $$ V=\pi\left[\frac{4}{3}y^{3/2}-\frac{y^2}{2}-\frac{y^3}{3}\right]_0^1 =\pi\left(\frac43-\frac12-\frac13\right) =\frac{\pi}{2}. $$
Check: at \(y=0\), both boundaries meet at \(x=0\), one unit from the axis, so the washer has zero area. For \(0\le y\le1\), \(\sqrt{y}\ge y\), hence \(R(y)\ge r(y)\). Differentiating the antiderivative gives \(2\sqrt{y}-y-y^2\), confirming the integration. The result is \(\frac{\pi}{2}\) cubic units.
Worked Example: The Axis Lies to the Right of the Region
Consider the region bounded by \(y=0\), \(y=1\), \(x=y^2\), and \(x=2\). Revolve it around \(x=3\) and find the volume.
The horizontal limits are \(0\le y\le1\). At each height the region extends from \(x=y^2\) to \(x=2\). The axis \(x=3\) is to the right of the region. Therefore the left endpoint is farther from the axis and gives the outer radius: $$ R(y)=3-y^2,\qquad r(y)=3-2=1. $$
The volume is $$ V=\pi\int_0^1\left((3-y^2)^2-1^2\right)\,dy =\pi\int_0^1(9-6y^2+y^4-1)\,dy =\pi\int_0^1(8-6y^2+y^4)\,dy. $$ An antiderivative is \(8y-2y^3+\frac{y^5}{5}\), so $$ V=\pi\left[8y-2y^3+\frac{y^5}{5}\right]_0^1 =\pi\left(8-2+\frac15\right) =\frac{31\pi}{5}. $$
Check: \(3-y^2\ge2\) on \(0\le y\le1\), so the outer radius is at least the inner radius \(1\), and the area is nonnegative. Differentiating the antiderivative gives \(8-6y^2+y^4\), the integrand used. The volume is \(\frac{31\pi}{5}\) cubic units.
A Common Radius Error
In the last example, a common mistake is to use the distance from the axis to the right endpoint as the outer radius. That would give \(R(y)=1\) and \(r(y)=3-y^2\), leading to $$ \pi\int_0^1\left(1-(3-y^2)^2\right)\,dy. $$ This has a negative integrand because \(3-y^2\ge2>1\). It cannot represent washer areas.
The correction is to compare the distances from both endpoints to \(x=3\). The left endpoint \(x=y^2\) is farther away, so \(R(y)=3-y^2\); the right endpoint \(x=2\) is nearer, so \(r(y)=1\). Always identify the axis location and compare distances, rather than assigning “outer” to a particular curve automatically.
What the Setup Depends On
A horizontal washer integral is appropriate when horizontal slices of the region, rotated about a vertical axis, produce annular cross-sections throughout the interval of integration. For a single pair of boundary curves, their \(x\)-coordinates must describe the slice correctly over the chosen \(y\)-range. If the active boundary changes or the region’s horizontal extent changes, the setup may need separate intervals.
The method is useful precisely because it matches the geometry of the rotation: every radius is measured perpendicular to the vertical axis, and therefore horizontally. Solving for inverse boundaries is not an algebraic formality; it supplies the endpoints needed to measure those radii. Before integrating, check the vertical limits, the selected inverse branches, the axis position, and the inequalities \(R(y)\ge r(y)\ge0\).
Check Your Understanding
For each question, identify the horizontal slice geometry before computing an integral.
- For the region between \(y=x^2\) and \(y=x\) on \(0\le x\le1\), what are the left and right \(x\)-boundaries at height \(y\)?
- If that region is revolved around \(x=2\), which boundary gives the outer radius, and why?
- For the region in the third worked example, explain why the radius from \(x=y^2\) is the outer radius when the axis is \(x=3\).
- What condition must hold between the outer and inner radii for a washer-area integrand to be nonnegative?
- When a curve has more than one solution for \(x\) at a given \(y\), what determines which solution should be used as a boundary?
