Volume of the tetrahedron x + y + z ≤ 1

∫₀¹∫₀¹⁻ˣ∫₀¹⁻ˣ⁻ʸ dz dy dx = 1/6 ≈ 0.16666667. A worked triple integral in Cartesian coordinates, with an editable calculator preloaded with the limits.

∫₀¹∫₀¹⁻ˣ∫₀¹⁻ˣ⁻ʸ dz dy dx = 1/6 ≈ 0.16666667. A textbook example of variable limits: each level depends on the ones outside it. The answer is 1/6.

Coordinate system

dV = dz dy dx

Integrand f

In terms of x, y, z

Limits of integration

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

x fromtoconstants only

y fromtomay depend on x

z fromtomay depend on x, y

Try an example

The setup

∫₀¹∫₀¹⁻ˣ∫₀¹⁻ˣ⁻ʸ dz dy dx = 1/6 ≈ 0.16666667

A textbook example of variable limits: each level depends on the ones outside it. The answer is 1/6.

Coordinate system: Cartesian. The integrand is 1, integrated from 0 to 1-x-y innermost, then 0 to 1-x, then 0 to 1.

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 ∫∫∫ (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 ∫₀¹∫₀¹⁻ˣ∫₀¹⁻ˣ⁻ʸ dz dy dx? ▾

It evaluates to 1/6 ≈ 0.16666667. A textbook example of variable limits: each level depends on the ones outside it. The answer is 1/6.

Why Cartesian coordinates for this one? ▾

The region is bounded by planes and graphs, so rectangular limits describe it directly with no change of variables needed.

Do I need to include the Jacobian myself? ▾

There is no Jacobian in Cartesian coordinates — dV is simply dz dy dx.

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