Volume cylinder calculus: how V = πr²h is derived
Volume cylinder calculus in one line: slice the cylinder into thin disks of area πr² and add them up the height, V = ∫₀ʰ πr² dz = πr²h. The same result comes without calculus, because a cylinder is a prism with a circular base, and every prism's volume is base area × height. Below, five derivations of the volume of a cylinder, from the simplest to the most formal.
Contents
V = πr²h. With r = 3 in and h = 8 in: base area π × 9 = 28.27 in², volume 28.27 × 8 = 226.19 in³.
The calculator above applies the result: give it a radius or diameter and a height and it returns the volume with each step. For the practical method, units and everyday examples, see the guide on how to find the volume of a cylinder.
1. Stacking disks: the intuitive derivation
Picture a stack of identical coins. Each coin is a very short cylinder: a disk of area πr² and thickness t, so its volume is πr² × t. Stack n coins and the height is h = n × t. The total volume is n × πr² × t = πr² × (n × t) = πr²h.
The argument works for any thickness, which is the point: the volume of a cylinder is its base area "repeated" over the height. This is also why the volume is proportional to the height (double h, double V) but to the square of the radius (double r, four times V).
2. Prisms with more and more sides
A prism's volume is base area × height. Inscribe a regular polygon with n sides in the circle of radius r and build a prism on it with height h. Its base area is (n ÷ 2) × r² × sin(2π ÷ n). As n grows, the polygon hugs the circle and its area tends to πr².
| Sides n | Inscribed polygon area (r = 1) | Circumscribed polygon area (r = 1) | Inscribed as % of π |
|---|---|---|---|
| 4 | 2.00000 | 4.00000 | 63.662 % |
| 6 | 2.59808 | 3.46410 | 82.699 % |
| 12 | 3.00000 | 3.21539 | 95.493 % |
| 24 | 3.10583 | 3.15966 | 98.862 % |
| 96 | 3.13935 | 3.14271 | 99.929 % |
The cylinder sits between the inscribed prism and the circumscribed prism (base area n × r² × tan(π ÷ n)). Both bounds close in on π × r² × h, so that is the cylinder's volume. Bounding a circle between polygons this way goes back to Archimedes; the 96-sided polygons in the last row are the ones he used to pin down π.
3. Cavalieri's principle
Cavalieri's principle says: if two solids have the same height and every horizontal slice of one has the same area as the slice of the other at the same level, the two solids have the same volume. Put a cylinder of radius r next to a box whose base is a square of area πr² (side r√π) and of the same height. Every slice of both is πr². The box's volume is πr² × h, so the cylinder's is too.
Cavalieri's principle also explains why a slanted cylinder holds as much as a straight one with the same base and vertical height: shifting the slices sideways does not change their areas. The oblique cylinder volume page uses exactly this.
4. The integral: V = ∫₀ʰ πr² dz
Calculus turns the stack of coins into an exact sum. Put the axis of the cylinder along z, from z = 0 to z = h. The cross-sectional area at height z is A(z). The volume of any solid is
V = ∫₀ʰ A(z) dz
For a cylinder, every slice is the same circle, so A(z) = πr² is a constant (this constant slice is the cross-sectional area of a cylinder). The integral of a constant is that constant times the length of the interval:
V = ∫₀ʰ πr² dz = πr² [z]₀ʰ = πr² (h − 0) = πr²h
Check with numbers: r = 3 in and h = 8 in give π × 9 × 8 = 226.19 in³.
The disk method (solid of revolution)
Rotate the line y = r, for x from 0 to h, about the x-axis. It sweeps out a cylinder of radius r and length h. The disk method gives
V = ∫₀ʰ π [f(x)]² dx = ∫₀ʰ π r² dx = πr²h
The shell method
Now rotate the rectangle 0 ≤ x ≤ r, 0 ≤ y ≤ h about the y-axis. Each thin vertical strip at distance x from the axis sweeps a cylindrical shell of circumference 2πx, height h and thickness dx, so volume 2πxh dx. Add the shells from the axis out to the edge:
V = ∫₀ʳ 2πx h dx = 2πh [x² ÷ 2]₀ʳ = πr²h
The same integral, from an inner radius r to an outer radius R, gives the volume of a tube: ∫ from r to R of 2πxh dx = πh(R² − r²). For R = 3 in, r = 2 in and h = 8 in that is 40π ≈ 125.66 in³; the hollow cylinder volume calculator does it for any size.
Triple integral in cylindrical coordinates
In cylindrical coordinates (ρ, θ, z) the volume element is ρ dρ dθ dz. Over the solid cylinder:
V = ∫₀ʰ ∫₀^2π ∫₀ʳ ρ dρ dθ dz = h × 2π × r² ÷ 2 = πr²h
5. The diameter and circumference forms
Everything above uses the radius. Substitute r = d ÷ 2 and the formula becomes V = π(d ÷ 2)²h = πd²h ÷ 4. With d = 6 in and h = 8 in, π × 36 × 8 ÷ 4 = 226.19 in³, the same as before. Substitute r = C ÷ (2π) and V = C²h ÷ (4π). The volume of a cylinder with the diameter page has worked examples of the d form.
Why dV/dr is the lateral area
Differentiate V = πr²h with respect to the radius, holding h fixed:
dV/dr = 2πrh
That is exactly the area of the curved side. The reason is geometric: increase the radius by a small dr and you wrap a thin shell around the cylinder, of volume about (side area) × dr = 2πrh × dr. For r = 3 in and h = 8 in, dV/dr = 48π ≈ 150.80 in², and growing r from 3 to 3.1 in adds 15.33 in³, close to the estimate 150.80 × 0.1 = 15.08 in³.
In the same way, dV/dh = πr²: adding a thin layer dh on top adds a disk of area πr². This is the link between the volume and the surface area of a cylinder, 2πr² + 2πrh, and the side alone, the lateral surface area. It also gives a calculus derivation of the side area itself: 2πrh is the integral ∫₀ʰ 2πr dz of the circumference over the height.
Calculus problems with the cylinder volume
Related rates: filling a tank
Water flows into an upright cylindrical tank of radius 2 ft at 3 ft³ per minute. How fast does the level rise? V = πr²h with r fixed, so dV/dt = πr² × dh/dt, and dh/dt = 3 ÷ (π × 2²) ≈ 0.239 ft per minute. In metric, a tank of radius 0.5 m filled at 10 L/min (0.01 m³/min) rises 0.01 ÷ (π × 0.25) ≈ 0.0127 m, about 1.27 cm, per minute.
Optimization: least material for a given volume
For a closed cylinder holding a volume V, the surface is A = 2πr² + 2πrh. Replace h by V ÷ (πr²): A(r) = 2πr² + 2V ÷ r. Set the derivative to zero: 4πr − 2V ÷ r² = 0, so V = 2πr³. Then h = V ÷ (πr²) = 2r: the most economical closed cylinder is as tall as it is wide. For 1 liter (1,000 cm³), r ≈ 5.42 cm, h ≈ 10.84 cm and A ≈ 553.58 cm². For 231 in³ (one US gallon), r ≈ 3.33 in, h ≈ 6.65 in and A ≈ 208.41 in².
Archimedes' sphere and cylinder
A sphere of radius r fits exactly inside a cylinder of radius r and height 2r. The cylinder's volume is πr² × 2r = 2πr³ and the sphere's is (4 ÷ 3)πr³, exactly two thirds. With r = 3 in: 169.65 in³ for the cylinder, 113.10 in³ for the sphere. Archimedes proved this ratio; the page on the cylinder, cone and sphere volumes compares all three.
Every derivation comes back to one idea: a cylinder's slices are all the same circle, so its volume is that circle's area times the height.
To use the result rather than prove it, enter your own measurements in the cylinder volume calculator.
FAQ
How do you derive the volume of a cylinder?
Cut the cylinder into thin disks of thickness dz. Each has volume πr² dz. Adding them from the bottom (z = 0) to the top (z = h) gives the integral ∫₀ʰ πr² dz = πr²h. Without calculus: a cylinder is a prism with a circular base, so its volume is base area × height.
How do you find the volume of a cylinder using integration?
Integrate the cross-sectional area along the axis: V = ∫₀ʰ A(z) dz. For a cylinder A(z) = πr² is constant, so V = πr² × h. The disk method, the shell method and a triple integral in cylindrical coordinates all give the same result.
Why is the derivative of the volume of a cylinder its lateral area?
dV/dr = d(πr²h)/dr = 2πrh. Growing the radius by a small amount dr adds a thin shell around the side, of volume about 2πrh × dr. So the rate at which volume grows with the radius is the area of the curved side.
What is the derivation of the formula V = πd²h/4?
Substitute r = d ÷ 2 into V = πr²h: π × (d ÷ 2)² × h = πd²h ÷ 4. It is the same formula written with the diameter.
How do you derive the volume of a hollow cylinder?
Subtract the inner cylinder from the outer one: πR²h − πr²h = π(R² − r²)h. With shells: ∫ from r to R of 2πxh dx = πh(R² − r²).
Related
- How to find the volumeHow to find the volume of a cylinder: square the radius, multiply by π and the height. Worked examples in inches, feet, cm and m, plus common mistakes.
- Oblique cylinderAn oblique cylinder has the same volume as a right one: V = πr²h with the perpendicular height. Use h = s × sin α from a slant side, with worked examples.
- Surface area of a cylinderThe surface area of a cylinder is A = 2πr² + 2πrh. Get the total, lateral and base areas from a radius or diameter, with worked examples in inches and cm.
Updated October 12, 2026