Disk Volumes About a Horizontal Axis

From a Rotated Region to Disks

When a plane region is rotated about a horizontal axis, a slice perpendicular to that axis sweeps out a circular cross section. If the region reaches the axis of rotation, the cross section is a solid disk: it has no hole in the middle.

This tutorial considers rotation about the x-axis. The key geometric measurement is each disk’s radius: the vertical distance from the x-axis to the boundary of the region. Once that radius is expressed as a function of $x$, the disk’s area can be integrated across the region.

Definition: If a region between $y=f(x)$ and the x-axis, for $a\le x\le b$, is rotated about the x-axis, and each vertical slice reaches the axis, its disk has radius $R(x)=|f(x)|$. If $f$ is continuous on $[a,b]$, the volume is $$ V=\pi\int_a^b [R(x)]^2\,dx =\pi\int_a^b [f(x)]^2\,dx. $$

The absolute value in the radius records that distance is nonnegative. Since squaring removes the sign, the volume integrand can also be written $f(x)^2$. The bounds $a$ and $b$ describe the horizontal extent of the region, so the integration variable is $x$.

Why the Disk Integral Works

A vertical slice at a fixed $x$ is a line segment from the x-axis to the curve. Rotating that segment about the x-axis sweeps out a disk. A disk of radius $R(x)$ has area $\pi[R(x)]^2$.

The slicing principle says that volume is obtained by accumulating cross-sectional area along the direction in which the slices are stacked. Here the slices have thickness $dx$, so a thin slice contributes approximately $\pi[R(x)]^2\,dx$ to the volume. The integral is the limit of sums of these contributions.

Choose a vertical slice.
At position $x$, the slice is perpendicular to the x-axis and has a vertical length equal to the distance from the curve to the axis.

Rotate the slice.
Because the segment reaches the x-axis, its rotation forms a disk of radius $R(x)$, with area $\pi[R(x)]^2$.

Accumulate the disk areas.
Adding cross-sectional areas over the interval gives $V=\int_a^b\pi[R(x)]^2\,dx$, provided the radius varies continuously.

More formally, partition $[a,b]$ into small intervals of widths $\Delta x_i$, and choose a sample point $x_i^*$ in each interval. The sum

\sum_i \pi[R(x_i^*)]^2\Delta x_i

approximates the volume by adding the volumes of thin cylinders. As the largest interval width approaches zero, these sums converge to the integral of the continuous cross-sectional area. Therefore,

V=\int_a^b \pi[R(x)]^2\,dx =\pi\int_a^b [R(x)]^2\,dx.

This derivation explains why the radius must be squared: it is the area of each circular cross section, not the circumference, that is being accumulated.

Worked Examples

Worked Example: A Curve Above the x-Axis

The region bounded by $y=\sqrt{x}$, the x-axis, $x=0$, and $x=4$ is rotated about the x-axis. Find its volume. Treat the coordinate units as centimeters.

At each $x$, the curve is above the axis, so the vertical distance to the axis is $R(x)=\sqrt{x}$. The bounds are $0$ and $4$. Thus

V=\pi\int_0^4(\sqrt{x})^2\,dx =\pi\int_0^4 x\,dx.

An antiderivative of $x$ is $x^2/2$, so

V=\pi\left[\frac{x^2}{2}\right]_0^4 =\pi\left(\frac{4^2}{2}-\frac{0^2}{2}\right) =\pi(8-0) =8\pi\ \text{cm}^3.

Check the antiderivative: $\frac{d}{dx}(x^2/2)=x$, the integrand. The volume is positive, as it must be.

Worked Example: A Curve Below the x-Axis

The region bounded by $y=-x-1$, the x-axis, $x=0$, and $x=2$ is rotated about the x-axis. Find the volume, with coordinates measured in centimeters.

The curve lies below the x-axis. The radius is its vertical distance from the axis: $R(x)=0-(-x-1)=x+1$. The distance is positive throughout the interval, and the disk-method integral is

V=\pi\int_0^2(x+1)^2\,dx =\pi\int_0^2(x^2+2x+1)\,dx.

An antiderivative is $x^3/3+x^2+x$. Evaluating at both bounds gives

\begin{aligned} V &=\pi\left[\frac{x^3}{3}+x^2+x\right]_0^2\\ &=\pi\left(\frac{8}{3}+4+2-0\right)\\ &=\pi\left(\frac{8}{3}+\frac{12}{3}+\frac{6}{3}\right)\\ &=\frac{26\pi}{3}\ \text{cm}^3. \end{aligned}

Check by differentiating: $\frac{d}{dx}(x^3/3+x^2+x)=x^2+2x+1=(x+1)^2$. The radius has been measured as a positive distance even though the graph is below the axis.

Worked Example: A Parabola Meeting the Axis at Both Bounds

The region bounded by $y=4-x^2$ and the x-axis, from $x=-2$ to $x=2$, is rotated about the x-axis. Find its volume.

On this interval, $4-x^2\ge 0$, so the radius is $R(x)=4-x^2$. The bounds are the two intersection points with the axis, found from $4-x^2=0$, which gives $x=-2$ and $x=2$. Therefore,

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

An antiderivative is $16x-\frac{8}{3}x^3+\frac{x^5}{5}$. Substituting the bounds,

\begin{aligned} V &=\pi\left[16x-\frac{8}{3}x^3+\frac{x^5}{5}\right]_{-2}^{2}\\ &=\pi\left[\left(32-\frac{64}{3}+\frac{32}{5}\right) -\left(-32+\frac{64}{3}-\frac{32}{5}\right)\right]\\ &=\pi\left(64-\frac{128}{3}+\frac{64}{5}\right)\\ &=\pi\left(\frac{960}{15}-\frac{640}{15}+\frac{192}{15}\right)\\ &=\frac{512\pi}{15}. \end{aligned}

The antiderivative checks because its derivative is $16-8x^2+x^4=(4-x^2)^2$. The integrand is nonnegative throughout the interval, and the computed volume is positive.

Conditions and a Reliable Setup

The basic disk formula applies when the rotated cross section is a full disk: the slice of the region reaches the axis of rotation. In these examples, the region lies between one curve and the x-axis. A region that leaves a gap around the axis would produce a cross section with a hole and would require a different setup; the disk formula alone would not describe that cross section.

Before integrating, identify the region and its horizontal bounds. Then express the radius as a vertical distance, square it to find the circular area, and integrate with respect to $x$. If the curve crosses the x-axis, the distance is $|f(x)|$; the volume integrand is still $f(x)^2$. The bounds must cover the entire region being rotated.

Key takeaway: For a region between $y=f(x)$ and the x-axis on $[a,b]$, rotated about the x-axis, use $$ R(x)=|f(x)|,\qquad V=\pi\int_a^b [R(x)]^2\,dx =\pi\int_a^b [f(x)]^2\,dx, $$ when each slice forms a full disk.

A Common Mistake: Forgetting to Square the Radius

For a disk, the cross-sectional area is $\pi R^2$, not $\pi R$. For example, for the line $y=4-x$ above the x-axis on $0\le x\le4$, the radius is $4-x$. The incorrect setup

V_{\text{wrong}}=\pi\int_0^4(4-x)\,dx =\pi\left[4x-\frac{x^2}{2}\right]_0^4 =8\pi

uses the radius where the disk’s area is required. It also has units of square length, not cubic length. The corrected calculation is

\begin{aligned} V &=\pi\int_0^4(4-x)^2\,dx\\ &=\pi\left[16x-4x^2+\frac{x^3}{3}\right]_0^4\\ &=\pi\left(64-64+\frac{64}{3}\right)\\ &=\frac{64\pi}{3}. \end{aligned}

Differentiating the antiderivative gives $16-8x+x^2=(4-x)^2$, confirming the integrand. The correction squares the radius before integrating, producing the required volume.

Why the Radius Is a Vertical Distance

The x-axis is horizontal, and the slices used here are vertical. At a chosen $x$, the radius runs vertically from the axis $y=0$ to the curve $y=f(x)$. Its length is therefore the difference in vertical coordinates, taken as a nonnegative distance: $|f(x)-0|$. This geometric interpretation helps distinguish the radius from the horizontal coordinate $x$, the slice thickness $dx$, and the curve’s signed height.

After integration, volume units follow from the cross-sectional area times the slice thickness: square length multiplied by length gives cubic length. Keeping track of this dimensional check is a useful way to notice an omitted square or an incorrect integration variable.

Check Your Understanding

For each question, identify the radius and the interval before evaluating or interpreting the integral.

A region between $y=3$ and the x-axis for $0\le x\le2$ is rotated about the x-axis. Write the volume integral.
For $y=-2x-1$ on $0\le x\le1$, what is the radius as a positive vertical distance from the x-axis?
A region between $y=x^2-1$ and the x-axis on $[-1,1]$ is rotated about the x-axis. Write a volume integral that accounts for the curve being below the axis.
Why must the radius be squared in the disk-method integral?
What units should a volume have if the coordinates are measured in centimeters?

Disk Volumes About a Horizontal Axis

What You'll Learn

From a Rotated Region to Disks

Why the Disk Integral Works