Disk Slices Around a Boundary Axis

From a Plane Region to a Solid

A plane region becomes a three-dimensional solid when it is revolved around a line in its plane. The volume can be found by examining thin slices of the region perpendicular to the axis of rotation. When each such slice reaches the axis and its rotation fills a disk, the solid’s cross-sectional area is the area of that disk.

The key modeling decisions are geometric: choose the slice direction, identify the distance from the axis to the farthest point of the slice, and determine the interval over which the region exists. The integral then adds the areas of these circular cross sections.

The radius is a distance, so it is nonnegative. It is not automatically the vertical height or horizontal width of the original region: it is the distance from the axis of rotation to the farthest point in the slice.

The Disk Integral

Suppose a region is revolved around the \(x\)-axis, and a vertical slice at position \(x\) extends from the axis \(y=0\) to a curve \(y=f(x)\), where \(f(x)\geq 0\) on \(a\leq x\leq b\). Rotating that slice fills a disk of radius \(R(x)=f(x)\). The cross-sectional area is therefore \(A(x)=\pi[f(x)]^2\).

By the slice principle, the volume is the integral of the cross-sectional area over the interval of slice positions:

The same idea works with a horizontal axis of rotation. If the axis is \(y=c\), and a vertical slice extends from that axis to a curve \(y=g(x)\) on one side of it, the radius is the distance \(|g(x)-c|\). Integrating with respect to \(x\) still uses disk areas, provided the rotated slices really are disks. For a vertical axis, the perpendicular slices are horizontal; the integral is then with respect to \(y\).

Why the Integral Gives the Volume

The formula follows from approximating the solid by thin slabs. For rotation around the \(x\)-axis, divide \([a,b]\) into many short intervals. In one interval, choose a sample point \(x_i^*\). If the interval has width \(\Delta x_i\), the slab there is approximately a disk of area \(\pi[R(x_i^*)]^2\) and thickness \(\Delta x_i\). Its approximate volume is the product of area and thickness.

Adding these approximations gives a Riemann sum. As the interval widths approach zero, the sums approach the volume, and the Riemann sum becomes the definite integral of the cross-sectional area:

This reasoning requires the cross-sectional area to be integrable over the interval; for the usual applications, a continuous radius function on a closed interval is sufficient. The square on the radius is essential: the area of each circular slice, not its circumference, contributes to the volume.

Worked Example: A Region Under a Square-Root Curve

Worked Example: A Linear Radius

Worked Example: Rotation Around a Vertical Boundary Axis

Reading the Radius and Bounds Carefully

Before integrating, describe one representative slice in words or with a sketch. Record its position, the endpoints of the slice, its distance from the axis, and the interval of positions over which the region exists. This separates the geometry from the calculus and helps prevent confusing the radius with some other length.

A slice that touches the axis often produces a disk, but the shape of the rotated cross section must be checked rather than inferred only from how the original region is described. If a slice consists of multiple segments, determine the distances of its points from the axis; the rotated cross section may or may not be a disk.

For instance, the slice \(y\in[-2,-1]\cup[0,1]\), rotated around the \(x\)-axis, produces an annulus with radii \(1\) and \(2\) from the first segment, and a disk of radius \(1\) from the second. Their union is a disk of radius \(2\). Thus multiple segments do not by themselves rule out a disk. The relevant question is whether the set of distances from the axis fills every radius from zero to the outer radius.

A Common Radius Error

Suppose the region under \(y=2x\) for \(0\leq x\leq3\) is revolved around the \(x\)-axis. A tempting but incorrect setup is to use \(x\) as the radius, perhaps because \(x\) is the slice position. That gives the wrong calculation:

The slice is vertical, and its distance from the \(x\)-axis to the curve is \(2x\), not \(x\). The correct radius is \(R(x)=2x\), so

To check, the antiderivative \(4\pi x^3/3\) differentiates to \(4\pi x^2=\pi(2x)^2\), and its endpoint difference is \(36\pi\). The error came from choosing a coordinate as the radius without measuring the distance to the axis.

Why the Disk Method Is Useful

The disk method converts a geometric solid into a one-variable integral. Its strength is that the cross-sectional area has a direct interpretation: every slice is a circular disk, and the area is determined by one radius. This makes the method especially effective when the region has a boundary on the axis of rotation and the other boundary gives a simple radius function.

In practical applications, units provide another check. If \(R\) is measured in units of length, then \(\pi R^2\) has units of area. Multiplying by a slice thickness and integrating across length produces cubic units, as a volume should. A result with square units usually signals that a slice thickness has been omitted.

For a reliable setup, confirm three things before evaluating: the slices are perpendicular to the axis, the rotated slices are disks, and the radius and bounds describe the whole region. Once these are correct, the volume is the definite integral of the disk areas.