Volume of a sphere in spherical coordinates

∫₀²ᵖⁱ∫₀ᵖⁱ∫₀² ρ² sin φ dρ dφ dθ = 32π/3 ≈ 33.51032164. A worked triple integral in spherical coordinates, with an editable calculator preloaded with the limits.

∫₀²ᵖⁱ∫₀ᵖⁱ∫₀² ρ² sin φ dρ dφ dθ = 32π/3 ≈ 33.51032164. Integrating 1 over a solid gives its volume — here 4πr³/3 with r = 2.

Coordinate system

dV = ρ² sin φ dρ dφ dθ — the Jacobian is added for you

Integrand f

In terms of θ, φ, ρ

Limits of integration

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

θ fromtoconstants only

φ fromtomay depend on θ

ρ fromtomay depend on θ, φ

Try an example

The setup

∫₀²ᵖⁱ∫₀ᵖⁱ∫₀² ρ² sin φ dρ dφ dθ = 32π/3 ≈ 33.51032164

Integrating 1 over a solid gives its volume — here 4πr³/3 with r = 2.

Coordinate system: spherical. The integrand is 1, integrated from 0 to 2 innermost, then 0 to pi, 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 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 Volume of a cone in cylindrical coordinates ∫₀²ᵖⁱ∫₀¹∫₀¹⁻ʳ r dz dr dθ

Frequently asked questions

What is the value of ∫₀²ᵖⁱ∫₀ᵖⁱ∫₀² ρ² sin φ dρ dφ dθ? ▾

It evaluates to 32π/3 ≈ 33.51032164. Integrating 1 over a solid gives its volume — here 4πr³/3 with r = 2.

Why spherical coordinates for this one? ▾

The region is measured naturally as a distance from the origin, so spherical coordinates turn a curved boundary into a constant limit on ρ.

Do I need to include the Jacobian myself? ▾

No. Enter only your integrand; the calculator multiplies by ρ² sin φ for you.

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