Washers Around the X-Axis

From a Region to a Solid

A plane region between two curves can be turned into a three-dimensional solid by rotating it around the x-axis. To calculate the solid’s volume, imagine cutting it into thin slices perpendicular to the x-axis. Each vertical slice sweeps out a circular ring, called a washer.

The essential setup question is which curve gives the outside edge of each ring and which gives the hole. For a region that lies above the x-axis, the upper curve is farther from the axis and determines the outer radius; the lower curve determines the inner radius. Subtracting the hole’s area from the full circular area gives the area of each washer.

Definition: Suppose $f$ and $g$ are continuous on $[a,b]$, and $0\leq g(x)\leq f(x)$ throughout that interval. Rotate the region between $y=f(x)$ and $y=g(x)$ about the x-axis. A vertical slice at $x$ forms a washer with outer radius $R(x)=f(x)$ and inner radius $r(x)=g(x)$.

This tutorial focuses on that setting: the region stays on or above the x-axis, and the same curve remains the upper boundary across the interval. These conditions make the radii and the integral direct to identify. If the region crosses the axis or the boundary ordering changes, the setup needs further analysis.

The Washer Integral Around the X-Axis

A circle of radius $R$ has area $\pi R^2$. Removing a concentric circle of radius $r$ leaves an annulus with area $\pi R^2-\pi r^2$. Therefore, at position $x$, the cross-sectional area of the washer is $\pi(f(x)^2-g(x)^2)$. Integrating these cross-sectional areas over the interval gives the volume.

Theorem: If $f$ and $g$ are continuous on $[a,b]$ and satisfy $0\leq g(x)\leq f(x)$ for every $x\in[a,b]$, then the volume of the solid formed by rotating the region between their graphs about the x-axis is $$ V=\pi\int_a^b\left(f(x)^2-g(x)^2\right)\,dx. $$ Here $f(x)$ is the outer radius and $g(x)$ is the inner radius.

The formula follows by adding the washer volumes across the interval. A thin slice of width $\Delta x$ near $x$ has approximate volume $\pi(f(x)^2-g(x)^2)\Delta x$. Summing such slices and taking the limit as their widths tend to zero gives the definite integral of their cross-sectional areas.

V=\int_a^b \pi\left(R(x)^2-r(x)^2\right)\,dx =\pi\int_a^b\left(R(x)^2-r(x)^2\right)\,dx.

The radii are lengths measured from the axis of rotation, not the thickness of the region. Because the axis here is $y=0$, a boundary at height $y\geq0$ is that distance $y$ from the axis. The slice thickness is instead measured horizontally, so it is represented by $dx$.

A Reliable Setup

Before integrating, describe the region as $x$ ranges over an interval and compare the two curve heights on that interval. In this setting, the larger height is the outer radius and the smaller height is the inner radius. Once the radii are identified, square each radius, subtract the inner squared radius from the outer squared radius, and integrate with respect to $x$.

Set the interval.
Find the left and right x-values that bound the region. They may come from intersections of the two curves or from specified vertical boundaries.

Compare the curve heights.
Determine which curve is above the other on the chosen interval. In this tutorial’s setting, both heights are nonnegative.

Name the radii.
The upper height is $R(x)$, the outer radius; the lower height is $r(x)$, the inner radius.

Integrate the washer area.
Use $V=\pi\int_a^b(R(x)^2-r(x)^2)\,dx$, then evaluate the definite integral.

It is worth writing down the radii before simplifying the integrand. Doing so makes the geometric meaning of each term visible and reduces the chance of subtracting in the wrong order.

Worked Examples

Worked Example: A Region Bounded by a Parabola and a Line

Find the volume formed by rotating the region between $y=4-x^2$ and $y=1$ about the x-axis.

First find where the curves meet:

4-x^2=1 \quad\Longrightarrow\quad x^2=3 \quad\Longrightarrow\quad x=-\sqrt{3}\ \text{or}\ x=\sqrt{3}.

Thus the region extends from $x=-\sqrt{3}$ to $x=\sqrt{3}$. On this interval $4-x^2\geq1$, so the parabola is the upper curve. The outer radius is $R(x)=4-x^2$, and the inner radius is $r(x)=1$. Both are nonnegative.

\begin{aligned} V &=\pi\int_{-\sqrt{3}}^{\sqrt{3}}\left((4-x^2)^2-1^2\right)\,dx\\ &=\pi\int_{-\sqrt{3}}^{\sqrt{3}}\left(16-8x^2+x^4-1\right)\,dx\\ &=\pi\int_{-\sqrt{3}}^{\sqrt{3}}\left(15-8x^2+x^4\right)\,dx. \end{aligned}

An antiderivative of the integrand is $15x-\frac{8x^3}{3}+\frac{x^5}{5}$. The integrand is even, so evaluate from $0$ to $\sqrt{3}$ and double:

\begin{aligned} V &=2\pi\left[15x-\frac{8x^3}{3}+\frac{x^5}{5}\right]_0^{\sqrt{3}}\\ &=2\pi\left(15\sqrt{3}-\frac{8(3\sqrt{3})}{3} +\frac{9\sqrt{3}}{5}\right)\\ &=2\pi\left(15\sqrt{3}-8\sqrt{3} +\frac{9\sqrt{3}}{5}\right)\\ &=2\pi\left(\frac{44\sqrt{3}}{5}\right) =\frac{88\pi\sqrt{3}}{5}. \end{aligned}

At both endpoints, $4-x^2=1$, so the inner and outer radii agree and the end washers have zero area, as expected. The volume is positive, and its units are cubic units if the coordinates are measured in units of length.

Worked Example: A Constant Outer Radius

Rotate the region between $y=3$ and $y=x+1$, for $0\leq x\leq2$, about the x-axis.

Throughout the interval, $x+1$ ranges from $1$ to $3$. Therefore $3\geq x+1\geq0$. The constant curve gives the outer radius $R(x)=3$, while the line gives the inner radius $r(x)=x+1$.

\begin{aligned} V &=\pi\int_0^2\left(3^2-(x+1)^2\right)\,dx\\ &=\pi\int_0^2\left(9-x^2-2x-1\right)\,dx\\ &=\pi\int_0^2\left(8-2x-x^2\right)\,dx. \end{aligned}

An antiderivative is $8x-x^2-\frac{x^3}{3}$. Hence

\begin{aligned} V &=\pi\left[8x-x^2-\frac{x^3}{3}\right]_0^2\\ &=\pi\left(16-4-\frac{8}{3}-0\right)\\ &=\frac{28\pi}{3}. \end{aligned}

As a check on the evaluation, integrate the unsimplified expression: an antiderivative of $9-(x+1)^2$ is $9x-\frac{(x+1)^3}{3}$. Its endpoint difference is $9-(-\frac13)=\frac{28}{3}$, agreeing with the result.

Worked Example: Two Nonconstant Radii

Rotate the region between $y=2+x$ and $y=1$, for $-1\leq x\leq1$, about the x-axis.

On this interval, $2+x$ ranges from $1$ to $3$. Thus the upper curve is $y=2+x$, and the lower curve is $y=1$. The radii are $R(x)=2+x$ and $r(x)=1$.

\begin{aligned} V &=\pi\int_{-1}^1\left((2+x)^2-1^2\right)\,dx\\ &=\pi\int_{-1}^1\left(4+4x+x^2-1\right)\,dx\\ &=\pi\int_{-1}^1\left(3+4x+x^2\right)\,dx. \end{aligned}

An antiderivative is $3x+2x^2+\frac{x^3}{3}$. Evaluating at both bounds gives

\begin{aligned} V &=\pi\left[3x+2x^2+\frac{x^3}{3}\right]_{-1}^1\\ &=\pi\left(\left(3+2+\frac13\right) -\left(-3+2-\frac13\right)\right)\\ &=\pi\left(\frac{16}{3}-\left(-\frac{4}{3}\right)\right) =\frac{20\pi}{3}. \end{aligned}

The $4x$ term contributes zero over the symmetric interval, since its antiderivative $2x^2$ has the same value at both endpoints. The remaining terms give a positive volume, consistent with the washer areas being nonnegative.

Why the Order of Subtraction Matters

The annulus area is the area of the outer disk minus the area of the inner disk. That geometric order is essential. For example, in the second worked example, reversing the radii would give

\pi\int_0^2\left((x+1)^2-3^2\right)\,dx =-\frac{28\pi}{3}.

This is not a volume: it is negative. The calculation has subtracted the larger disk from the smaller one. On the whole interval, $3\geq x+1$, so the correct order is $3^2-(x+1)^2$. The corrected integral evaluates to $\frac{28\pi}{3}$.

Common mistake: Do not decide which radius is outer from the names or appearance of the curve equations. Compare their heights on the interval. For a region above the x-axis, the larger height gives the outer radius, and its square must come first in $R(x)^2-r(x)^2$.

Interpreting and Checking the Result

A washer integral combines two-dimensional cross-sectional areas with a one-dimensional interval of integration. Each squared radius has units of length squared, and $dx$ has units of length, so the resulting volume has units of length cubed. The factor $\pi$ is dimensionless.

Several checks are useful after setting up an integral:

Confirm that the interval describes the full region being rotated.
Check that the proposed outer radius is at least as large as the inner radius throughout the interval.
Check that the integrand $\pi(R(x)^2-r(x)^2)$ is nonnegative.
Evaluate the antiderivative at the upper bound and subtract its value at the lower bound.

A zero inner radius is allowed: the cross-section at that $x$ is a full disk, which is the limiting case of a washer with no hole. But when the region remains a positive distance above the axis, the inner radius is positive and the cross-section has a genuine hole.

The central geometric idea is consistent throughout: take a vertical slice, measure both boundary heights from the x-axis, subtract the inner circular area from the outer circular area, and integrate over $x$. Correctly identifying those two distances is the key step in a washer setup.

Check Your Understanding

For each question, identify the geometry before carrying out any integration.

For the region between $y=5$ and $y=2$ on $0\leq x\leq1$, what are the outer and inner radii when it is rotated about the x-axis?
Write the washer integral for the region between $y=4-x$ and $y=1$ on $0\leq x\leq3$, rotated around the x-axis. Which curve gives the outer radius?
Why does the washer area use $R(x)^2-r(x)^2$, rather than $R(x)-r(x)$?
What geometric or arithmetic warning is indicated if a proposed washer integral has a negative integrand throughout its interval?
For the region between $y=3-x$ and $y=1$ on $0\leq x\leq2$, identify the radii and set up the volume integral.

What You'll Learn

From a Region to a Solid

The Washer Integral Around the X-Axis

A Reliable Setup

Worked Examples