Cylinder Volume Using Integration

Cylinder Volume Using Integration is a calculus-based approach to deriving the cylinder volume formula. This page exists because integration is the mathematical foundation for computing volumes of all solids of revolution — and the cylinder is the cleanest, most accessible example of the disk method in action.

Using integration, we compute V = ∫₀ʰ πr² dy = πr²h by summing infinitely many infinitesimally thin circular disks stacked along the height axis. This proves the formula rigorously using calculus, rather than relying on geometric intuition alone.

This is essential for calculus students learning the disk and shell methods, AP/IB math preparation, and physics students who need to understand how volume integrals work before tackling more complex shapes like cones, spheres, and solids of revolution.

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.

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.