Cylinder Volume Using Integration
This page shows how to derive cylinder volume using calculus. The disk method integrates circular cross-sections along the height axis: V = ∫₀ʰ πr² dy = πr²h. If you're studying calculus, this is one of the cleanest examples of the disk method.
Volume Using Integration
The Disk Method
Place the cylinder with its base on the xy-plane and its axis along the y-axis, running from y = 0 to y = h. At any height y, the cross-section is a disk — a filled circle of radius r.
The area of each disk is A(y) = πr². Since r doesn't change with y, the volume integral is:
V = ∫₀ʰ πr² dy = πr² ∫₀ʰ dy = πr² [y]₀ʰ = πr²h
Result: V = πr²h. The integral confirms the geometric intuition.
Shell Method Alternative
You can also use the cylindrical shell method. Imagine the cylinder as nested cylindrical shells from x = 0 to x = r, each with height h.
Each shell at radius x has circumference 2πx, height h, and thickness dx. Its volume element is dV = 2πxh dx.
V = ∫₀ʳ 2πxh dx = 2πh [x²/2]₀ʳ = 2πh × r²/2 = πr²h
Both methods — disk and shell — give the same answer, as expected.
Extending to Variable Radius
If the radius changes along the height — like a tapered cylinder or a general solid of revolution — the integral adapts:
V = ∫₀ʰ π[r(y)]² dy
For a cone with radius that decreases linearly from R to 0: r(y) = R(1 − y/h). The integral gives V = πR²h/3 — one-third of the cylinder.
For a sphere of radius R centered at y = 0: r(y) = √(R² − y²). Integrating from −R to R gives V = (4/3)πR³. Integration handles any cross-section shape.