Disks Rotated Around the X-Axis

From a Bounded Region to a Volume

A curve and the x-axis can enclose a region. Rotating that region around the x-axis sweeps out a three-dimensional solid. To find its volume, examine a thin vertical slice of the region: as it rotates, the slice traces a filled circular disk.

The disk’s radius is the vertical distance from the curve to the x-axis. Its area therefore depends on the square of the curve’s height, whether the curve lies above or below the axis. The main tasks are to find the correct x-bounds, identify that distance, and integrate the disk areas across the region.

Finding the Bounds and the Radius

The bounds come from the intersections of the curve with the x-axis. Since points on the x-axis have \(y=0\), find these intersections by solving \(f(x)=0\). For a single bounded region between the curve and the axis, the relevant intersections provide the endpoints \(a\) and \(b\).

At each \(x\) between the endpoints, the slice runs from \(y=0\) to \(y=f(x)\), or from \(y=f(x)\) to \(y=0\) if the curve is below the axis. Rotating that segment around the x-axis fills a disk. The radius is a distance, so it is nonnegative: \(r(x)=|f(x)|\). But the disk area uses \(r(x)^2\), and \(|f(x)|^2=[f(x)]^2\). Thus the same integral formula works whether the curve is above or below the axis throughout the region.

Here is the setup in sequence:

Why the Disk Integral Gives Volume

For each \(x\) in the interval, the rotated cross-section is a disk of radius \(|f(x)|\). A disk of radius \(r\) has area \(\pi r^2\), so the cross-sectional area at \(x\) is \(A(x)=\pi[f(x)]^2\).

To see why integrating these areas gives the volume, divide \([a,b]\) into short subintervals. On a subinterval of width \(\Delta x\), the solid’s volume is approximately the area of a representative disk times \(\Delta x\). Adding these approximations gives a sum of the form \(\sum A(x_i^*)\Delta x\). As the widths shrink toward zero, these sums approach the definite integral of the cross-sectional area. Therefore, by the volume formula for slices, the volume is \(\int_a^b A(x)\,dx\). Substituting the disk area gives the disk integral.

This argument also explains why the curve’s sign does not make the volume negative: a cross-sectional area is nonnegative, and squaring the height produces a nonnegative integrand.

Worked Examples

Common Pitfall: Using the Signed Height as the Radius

A radius is a distance, so it cannot be negative. For a curve below the x-axis, writing \(r(x)=f(x)\) would produce a negative radius. In the second example, that mistake would give \(r(x)=x^2-1<0\) between \(-1\) and \(1\). The correction is to use the distance \(r(x)=|f(x)|=1-x^2\).

In this particular disk formula, the sign error can appear to disappear after squaring:

That identity explains why the volume integrand can be written as \(\pi[f(x)]^2\), even when the graph is below the axis. It does not change the geometric meaning of the radius: describe it as a nonnegative distance, then square it to find the area.

Choosing the Right Interval

For a bounded region between one curve and the x-axis, the endpoints are typically consecutive x-intercepts that enclose the region being rotated. If the equation \(f(x)=0\) has several solutions, use the intersections corresponding to the particular bounded region in the question. The interval matters: integrating across unrelated intercepts can include additional regions and produce a different solid.

Before evaluating, check that the setup matches the geometry:

• The interval endpoints are the x-axis intersections that bound the intended region.
• A vertical slice perpendicular to the axis of rotation creates a filled disk.
• The radius is the distance from the curve to the x-axis.
• The integrand is the disk area, \(\pi[f(x)]^2\), and the variable is \(x\).

Once these choices are correct, evaluate the definite integral and check the antiderivative and endpoint substitution. These checks catch common algebra and sign errors before they affect the final volume.