From Circular Slices to Volume
A solid formed by rotating a planar region around a straight axis can often be understood by examining its cross-sections perpendicular to that axis. When each such cross-section is a filled circle, or disk, the volume can be found by integrating the area of those disks along the axis.
The key geometric fact is that the area of a disk depends on the square of its radius. The radius is a length measured from the axis of rotation to the boundary of the region. As the slice moves along the axis, that distance may change, so the disk areas may change as well.
We will use a coordinate \(t\) measured along the axis of rotation. Let \(R(t)\) denote the radius of the circular slice at position \(t\). The disk formula expresses the volume in terms of this radius function.
Why the Radius Is Squared
At each fixed position \(t\), the slice is a circle filled in to its center, not just a circumference. A circle of radius \(R(t)\) has area \(\pi[R(t)]^2\). The square appears because area measures two perpendicular dimensions: when the radius is scaled by a factor, both dimensions of the disk scale by that factor.
To build up the solid, divide the interval \([a,b]\) into short pieces. Over a small piece of length \(\Delta t_i\), the radius is approximately constant, so the part of the solid there is approximately a short cylinder. Its approximate volume is the disk area multiplied by the slice thickness:
where \(t_i^*\) is a sample point in the \(i\)-th piece. Adding these approximate slice volumes gives a Riemann sum. As the pieces become thinner, this sum approaches the exact volume. This is the cross-sectional area principle: volume is the integral of the area of slices perpendicular to the direction of accumulation.
This derivation shows why the formula has this particular form. The factor \(\pi[R(t)]^2\) is the area of one circular slice; \(dt\) represents accumulation along the axis. The formula is not an integral of circumference, nor is it an integral of radius alone.
Reading the Radius from the Geometry
The radius is a distance, so it is nonnegative. At each location along the axis, measure from the axis itself to the boundary reached by the region. If the axis is described by one coordinate value and the boundary by another, the radius is the distance between those values. It is not automatically equal to the boundary's coordinate value: that is true only when the axis lies at coordinate zero and the coordinate measures perpendicular distance from it.
The disk formula applies when the rotated region reaches the axis at every position in the interval. A cross-section then fills from the center to its outer edge. If a region leaves a gap around the axis, the cross-section is not a disk; the disk formula in this form does not describe that slice. Here we consider only filled circular slices.
Worked Examples
Worked Example: A Region with Constant Radius
Take a rectangular region of length \(4\) units along an axis and width \(3\) units perpendicular to it. Rotate the rectangle around one of its long edges. At every position along the axis, the cross-section is a disk of radius \(3\) units. Find the volume.
The interval along the axis is \(0\le t\le 4\), and the radius is \(R(t)=3\) units. Substitute into the disk formula:
Check: every slice has area \(9\pi\) square units, and the solid has length \(4\) units. Thus its volume is \(9\pi\cdot 4=36\pi\) cubic units, agreeing with the integral.
Worked Example: A Radius That Grows Linearly
Consider a triangular region whose edge lies along an axis, with the opposite edge rising from radius \(0\) at \(t=0\) to radius \(2\) units at \(t=2\). Rotate the region around the axis. Find the volume.
The sloping edge has radius \(R(t)=t\) units for \(0\le t\le 2\). Each perpendicular slice is a disk, so
Check the integration: the derivative of \(\pi t^3/3\) is \(\pi t^2\), the integrand. At the endpoints, the disk areas are \(0\) and \(4\pi\) square units, consistent with the radius growing from \(0\) to \(2\). The volume is positive, as expected.
Worked Example: A Radius with a Positive Starting Value
Suppose a trapezoidal region lies between an axis and a straight boundary whose perpendicular distance from the axis is \(R(t)=1+t\) units for \(0\le t\le 1\). Rotate the region around the axis and find the volume.
The radius changes from \(R(0)=1\) unit to \(R(1)=2\) units. The disk area at \(t\) is \(\pi(1+t)^2\), so
Check: the derivative of \(\pi(t+t^2+t^3/3)\) is \(\pi(1+2t+t^2)=\pi(1+t)^2\). The radii at the two endpoints are \(1\) and \(2\) units, so the endpoint disk areas are \(\pi\) and \(4\pi\) square units. The calculated volume is positive and is consistent with disks whose radii grow across the interval.
A Complete View of the Derivation
The disk method follows from three precise steps. First, determine the interval along the axis that the solid occupies. Second, at each position in that interval, determine the perpendicular distance from the axis to the edge of the rotated region. Third, use the area of the resulting disk and accumulate those areas along the axis.
Choose a position \(t\) along the axis. The slice is taken in a plane perpendicular to the axis.
Find \(R(t)\), the nonnegative perpendicular distance from the axis to the boundary of the region.
The slice is a disk, so its area is \(\pi[R(t)]^2\).
Accumulate the cross-sectional areas over the interval to obtain volume.
In the Riemann-sum derivation, each term approximates the volume of a thin piece of the solid. The approximation improves as the slice thickness decreases. The integral is the limit of these sums, and the continuity or piecewise continuity of the radius ensures that the cross-sectional area can be integrated over the interval.
A Common Mistake: Integrating the Radius
A tempting but incorrect calculation is to integrate the radius itself. For the linearly growing radius in the second example, this would give
This is not the disk-method volume. The radius \(t\) has units of length, so integrating it along an interval produces units of area, not volume. More importantly, \(t\) is not the cross-sectional area. The disk area is \(\pi t^2\), so the correct calculation is
The correction is to square the radius when computing the area of each circular slice, then integrate that area along the axis.
Why the Disk Method Is Useful
The method turns a three-dimensional volume problem into a one-variable integral. Instead of trying to describe the entire solid at once, it uses a sequence of simple cross-sections whose areas can be determined from the geometry of the original region.
The central distinction is between the dimensions of a slice and its position in the solid. The radius is measured perpendicular to the axis; the variable \(t\) moves along the axis. Keeping those roles separate makes both the setup and the units easier to check. The disk method is appropriate when the slices are filled circles, and its integrand is the area of each such circle.
Check Your Understanding
Use the cross-sectional interpretation to answer these questions.
- What geometric quantity does \(R(t)\) represent in the disk formula?
- Why is the area of a slice \(\pi[R(t)]^2\), rather than \(\pi R(t)\)?
- A solid has disk radius \(4\) units at every position along an axis interval of length \(2\) units. Write the volume integral.
- For a radius \(R(t)=t+1\) on \(0\le t\le 2\), write the correct volume integrand.
- What feature of the rotated region distinguishes a disk slice from a circular slice with an unfilled gap at the center?
