Volume of a cone in cylindrical coordinates

∫₀²ᵖⁱ∫₀¹∫₀¹⁻ʳ r dz dr dθ = π/3 ≈ 1.04719755. A worked triple integral in cylindrical coordinates, with an editable calculator preloaded with the limits.

∫₀²ᵖⁱ∫₀¹∫₀¹⁻ʳ r dz dr dθ = π/3 ≈ 1.04719755. The height limit shrinks as r grows, tracing the slanted side of the cone. The answer is π/3.

Coordinate system

dV = r dz dr dθ — the extra r is added for you

Integrand f

In terms of θ, r, z

Limits of integration

Inner limits may use the outer variables — e.g. 1-x, sqrt(1-x^2)

θ fromtoconstants only

r fromtomay depend on θ

z fromtomay depend on θ, r

Try an example

The setup

∫₀²ᵖⁱ∫₀¹∫₀¹⁻ʳ r dz dr dθ = π/3 ≈ 1.04719755

The height limit shrinks as r grows, tracing the slanted side of the cone. The answer is π/3.

Coordinate system: cylindrical. The integrand is 1, integrated from 0 to 1-r innermost, then 0 to 1, then 0 to 2*pi.

More worked triple integrals

Triple integral over the unit cube ∫₀¹∫₀¹∫₀¹ xyz dz dy dx Volume of a sphere in spherical coordinates ∫₀²ᵖⁱ∫₀ᵖⁱ∫₀² ρ² sin φ dρ dφ dθ Volume of a cylinder in cylindrical coordinates ∫₀²ᵖⁱ∫₀¹∫₀³ r dz dr dθ Volume of the tetrahedron x + y + z ≤ 1 ∫₀¹∫₀¹⁻ˣ∫₀¹⁻ˣ⁻ʸ dz dy dx ∫∫∫ (x² + y² + z²) over the unit cube ∫₀¹∫₀¹∫₀¹ (x² + y² + z²) dz dy dx Mass of a solid with density ρ = z ∫₀²∫₀²∫₀² z dz dy dx Volume of a hemisphere ∫₀²ᵖⁱ∫₀^(π/2)∫₀¹ ρ² sin φ dρ dφ dθ

Frequently asked questions

What is the value of ∫₀²ᵖⁱ∫₀¹∫₀¹⁻ʳ r dz dr dθ? ▾

It evaluates to π/3 ≈ 1.04719755. The height limit shrinks as r grows, tracing the slanted side of the cone. The answer is π/3.

Why cylindrical coordinates for this one? ▾

The region is symmetric about the z-axis, so circular cross-sections become the simple limits 0 ≤ r ≤ R once you switch to cylindrical coordinates.

Do I need to include the Jacobian myself? ▾

No. Enter only your integrand; the calculator multiplies by r for you.

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