Disk Method for the Volume of a Solid of Revolution

A solid of revolution is formed by rotating a curved segment about an axis. A typical problem in calculus is to find the volume of the solid formed by rotating a function y = f(x) around either the y- or x-axis. If the function is revolved around the x-axis, you can find the volume using disk method.

The disk method divides the solid into circular cross-sections (disks) of infinitesimal thickness. The sum of the volumes of these disks is the volume of the solid. The infinite sum of infinitesimally thin disks can be calculated with integrals.

The volume integral formula for a curve rotated around the x-axis is

V = ∫πf(x)² dx.

How is this equation derived? First, imagine a solid of rotation and its circular cross-section at an arbitrary point x. Now imagine the very thin disk between x and x + Δx. The radius of this disk is the height of the function at x, so the height is f(x). The thickness of the disk is x + Δx - x = Δx. Thus, the volume of the disk is

π times f(x)² times Δx

As Δx becomes smaller, the sum of these disks approaches the true volume of the solid.

From geometry, we know that the volume of a cone is (π/3)r²h. But we consider a cone to be the solid that results from rotating a line about the x-axis, we can also find its volume using integral calculus.

For instance, consider the line y = 1 - x and consider the segment of this line between x = 0 and x = 1. This segment forms a triangle in the first quadrant. If we rotate this line segment about the x-axis, it forms a cone with a height of 1 and radius of 1. Using the geometry formula, we compute that its volume is π/3.

Using calculus, we set up the integral ∫π(1-x)² dx evaluated from 0 to 1.

∫π(1-x)² dx = ∫π(1 - 2x + x²) dx

= π[x - x² + (1/3)x³]

= π[1 - 1² + (1/3)1³] - π[0 - 0² + (1/3)0³]

= π/3

So as you can see the calculus formula matches the geometry formula.