Rotating a 2D region around an axis sweeps it through space and creates a 3D solid. Computing the volume of that solid uses the same idea as every other integral: slice it into thin pieces whose volumes you know, then integrate. For rotation, those thin pieces are disks or cylindrical shells, depending on how you orient the slice.

The Big Idea

Rotate the region between y = f(x) and the x-axis around the x-axis. The result is a solid of revolution. Slice it into thin disks, each with volume π[f(x)]² dx. Integrate to get the total.

The Disk Method

V = π ∫ab [f(x)]² dx
📋 Rotate y = √x on [0,4] around x-axis
SetupV = π∫₀⁴ (√x)² dx = π∫₀⁴ x dx
Integrate= π[x²/2]₀⁴ = π·8 = 8π

The Washer Method

For the region between two curves f(x) (outer) and g(x) (inner):

V = π ∫ab ([f(x)]² − [g(x)]²) dx

The Shell Method

Use cylindrical shells when rotating around the y-axis:

V = 2π ∫ab x · f(x) dx
Disk vs Shell

Disk: slice perpendicular to axis of rotation. Shell: slice parallel to axis. Both give the same answer — choose whichever integral is simpler to set up.

Frequently Asked Questions

How do I choose which method?▾
Rotating around x-axis with y = f(x): disk/washer is usually simpler. Rotating around the y-axis with x values: shells are usually simpler. Try both and see which integral looks cleaner.
Can I rotate around y = k?▾
Yes — rotating around y = k shifts the radius. The disk radius becomes f(x) − k (or k − f(x) if k is above the curve).

Why Slicing Works

The key insight is the same one that underlies Riemann sums: if you cut a 3D solid into thin enough slices, each slice is approximately a simple shape (a disk or a cylindrical shell) whose volume you can compute exactly. Add up all the slices and take the limit as they become infinitely thin — that is the integral.

Example — Volume of a Cone

Verify the formula V = (1/3)πr²h using integration. A cone of radius r and height h can be generated by rotating y = (r/h)x around the x-axis from x = 0 to x = h.

📋 Cone by disk method
SetupV = π∫₀ʰ [(r/h)x]² dx = π(r/h)² ∫₀ʰ x² dx
Integrate= π(r/h)² · [x³/3]₀ʰ = π(r/h)² · h³/3
Simplify= πr²h/3 = (1/3)πr²h ✓

Example — Volume of a Sphere

A sphere of radius r is generated by rotating y = √(r²−x²) around the x-axis from x = −r to x = r.

📋 Sphere by disk method
SetupV = π∫₋ᵣʳ (r²−x²) dx
Integrate= π[r²x − x³/3]₋ᵣʳ
Evaluate= π[(r³ − r³/3) − (−r³ + r³/3)] = π · (4r³/3) = (4/3)πr³ ✓

Setting Up the Integral — The Most Important Step

The most common source of errors is setting up the integral incorrectly. Before writing any integral, ask three questions:

📋 Setup checklist
1What axis am I rotating around? (determines whether you integrate over x or y)
2What is the radius of each disk/shell? (the function value, possibly adjusted for axis shifts)
3If there is a hole (washer), what is the inner radius?

Pappus's Theorem — A Useful Shortcut

For a region with area A whose centroid is at distance d from the axis of rotation, the volume of revolution is V = 2πd·A. This can save significant computation when you know the centroid. For example, rotating a circle of radius r whose centre is at distance R from the axis gives a torus with volume V = 2π · R · πr² = 2π²Rr².

AM
Dr. Aisha Malik, PhD Mathematics
Senior Lecturer in Applied Mathematics · 12 years teaching calculus
Dr. Malik holds a PhD in Applied Mathematics from the University of Edinburgh and has taught calculus to over 4,000 students. She has reviewed this article for mathematical accuracy and pedagogical clarity.
Technically reviewed by: Prof. James Chen, Stanford Mathematics Department · April 2026