Cylindrical Shells Method for the Volume of a Solid of Revolution

A solid of revolution is formed by spinning a curve around an axis. A typical calculus problem is to find the volume of the solid formed by rotating a function y = f(x) around either axis. If the function is revolved around the y-axis, you can find the volume using the method of cylindrical shells.

The cylindrical shell method partitions the solid into nested hollow cylinders (pipes) of infinitesimal thickness. The sum of the volumes of these shells gives the the volume of the solid. The infinite sum of infinitesimally thin shells can be calculated with integrals.

The volume of a function f(x) rotated around the y-axis is

V = ∫2πx*f(x) dx

Why does this formula express the volume? First, consider solid of rotation to be an infinite collection of nested hollow cylinders of infinitesimal thickness. Now consider the shell between an arbitrary point x and x + Δx. The volume of this shell is approximately equal to its height times its circumference times its thickness. The thinner the shell, the more accurate this formula is.

The height is f(x), the circumference is 2πx, and the thickness is Δx. When you multiply these elements together and express the sum in calculus notation, you end up with ∫2πx*f(x) dx.

We can use the cylindrical shells method to find the volume of a cone. For example, take the cone defined by rotating the line y = 1 - x around the y-axis, where the bounds are x = 0 and x = 1. This is a cone with a height of 1 and a radius of 1. If we use the simple geometry formula for the volume of a cone, (π/3)r²h, we can see its volume is π/3.

Does that match the answer obtained with calculus?

The integral to evaluate is ∫2πx(1-x) dx from 0 to 1. Doing the math, we have:

∫2πx(1-x) dx = ∫2π(x - x²) dx

= πx² - (2/3)πx³

= π[1² - (2/3)1³] - π[0² - (2/3)0³]

= π/3

So the values do match.