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Triple Integral Calculator

Evaluate triple integrals in rectangular, cylindrical, or spherical coordinates with AI-powered step-by-step solutions

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triple integral of xyz over [0,1]x[0,1]x[0,1]
triple integral of x^2+y^2+z^2 in spherical coords over unit ball
triple integral of z over cylinder x^2+y^2<=1, 0<=z<=2
triple integral of 1 over tetrahedron bounded by x+y+z=1 and axes

What is a Triple Integral?

A triple integral extends the concept of single and double integrals to three dimensions. For a function f(x,y,z)f(x, y, z) defined on a solid region E⊂R3E \subset \mathbb{R}^3:

∭Ef(x,y,z) dV\iiint_E f(x,y,z)\,dV

gives the total accumulation of ff over EE. The infinitesimal volume element dVdV becomes dx dy dzdx\,dy\,dz in Cartesian coordinates, but can be rewritten depending on the geometry of EE.

Common physical meanings:

  • If f(x,y,z)=1f(x,y,z) = 1, the integral gives the volume of EE.
  • If f(x,y,z)=ρ(x,y,z)f(x,y,z) = \rho(x,y,z) is a density, it gives the total mass.
  • Moments, centers of mass, and moments of inertia are all triple integrals of weighted density functions.

The key to evaluating a triple integral is choosing the right coordinate system and setting up the bounds correctly.

How to Set Up and Evaluate Triple Integrals

Step 1: Choose Coordinates

Region GeometryBest CoordinatesVolume Element
Box / generalRectangular (x,y,z)(x,y,z)dx dy dzdx\,dy\,dz
Cylindrical symmetryCylindrical (r,θ,z)(r, \theta, z)r dr dθ dzr\,dr\,d\theta\,dz
Spherical symmetrySpherical (ρ,φ,θ)(\rho, \varphi, \theta)ρ2sin⁡φ dρ dφ dθ\rho^2 \sin\varphi\,d\rho\,d\varphi\,d\theta

Step 2: Set Up Bounds

Project the region onto a coordinate plane to determine the order of integration. For type-I solid bounded above by z=g2(x,y)z = g_2(x,y) and below by z=g1(x,y)z = g_1(x,y):

∭Ef dV=∬D[∫g1(x,y)g2(x,y)f(x,y,z) dz]dA\iiint_E f \, dV = \iint_D \left[\int_{g_1(x,y)}^{g_2(x,y)} f(x,y,z)\,dz\right] dA

Step 3: Evaluate Iteratively

Integrate innermost first, treating outer variables as constants. Then proceed outward.

Cylindrical Coordinates

Use the substitutions x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, z=zz = z:

∭Ef(x,y,z) dV=∭Ef(rcos⁡θ,rsin⁡θ,z)⋅r dr dθ dz\iiint_E f(x,y,z)\,dV = \iiint_E f(r\cos\theta, r\sin\theta, z) \cdot r\,dr\,d\theta\,dz

The extra factor of rr comes from the Jacobian determinant.

Spherical Coordinates

Use x=ρsin⁡φcos⁡θx = \rho\sin\varphi\cos\theta, y=ρsin⁡φsin⁡θy = \rho\sin\varphi\sin\theta, z=ρcos⁡φz = \rho\cos\varphi:

∭Ef dV=∭Ef⋅ρ2sin⁡φ dρ dφ dθ\iiint_E f\,dV = \iiint_E f \cdot \rho^2 \sin\varphi\,d\rho\,d\varphi\,d\theta

The Jacobian ρ2sin⁡φ\rho^2 \sin\varphi is critical — forgetting it is the single most common error.

Common Mistakes to Avoid

  • Forgetting the Jacobian: Cylindrical gets a factor of rr, spherical gets ρ2sin⁡φ\rho^2 \sin\varphi. Skipping this gives a wrong answer every time.
  • Wrong bounds order: The innermost bounds may depend on outer variables, but the outermost bounds must be constants. Reversing this generates nonsense.
  • Sign errors with sin⁡φ\sin\varphi: In spherical, φ∈[0,π]\varphi \in [0, \pi] (so sin⁡φ≥0\sin\varphi \geq 0). Using φ∈[0,2π]\varphi \in [0, 2\pi] is wrong.
  • Mixing conventions: Some books use φ\varphi for the polar angle (from z-axis), others for the azimuthal angle. Be consistent with one convention.
  • Not sketching the region: For non-trivial solids, a quick sketch saves you from impossible bounds.

Examples

Step 1: Set up iterated integral: ∫01∫01∫01xyz dz dy dx\int_0^1 \int_0^1 \int_0^1 xyz \, dz\, dy\, dx
Step 2: Integrate over zz: ∫01xyz dz=xyz22∣01=xy2\int_0^1 xyz \, dz = \frac{xy z^2}{2}\big|_0^1 = \frac{xy}{2}
Step 3: Integrate over yy: ∫01xy2 dy=xy24∣01=x4\int_0^1 \frac{xy}{2} \, dy = \frac{x y^2}{4}\big|_0^1 = \frac{x}{4}
Step 4: Integrate over xx: ∫01x4 dx=x28∣01=18\int_0^1 \frac{x}{4} \, dx = \frac{x^2}{8}\big|_0^1 = \frac{1}{8}
Answer: 18\dfrac{1}{8}

Step 1: In spherical: 0≤ρ≤10 \leq \rho \leq 1, 0≤φ≤π0 \leq \varphi \leq \pi, 0≤θ≤2π0 \leq \theta \leq 2\pi
Step 2: Volume = ∫02π∫0π∫01ρ2sin⁡φ dρ dφ dθ\int_0^{2\pi} \int_0^\pi \int_0^1 \rho^2 \sin\varphi \, d\rho \, d\varphi \, d\theta
Step 3: Inner: ∫01ρ2 dρ=13\int_0^1 \rho^2 \, d\rho = \frac{1}{3}
Step 4: Middle: ∫0πsin⁡φ dφ=2\int_0^\pi \sin\varphi \, d\varphi = 2
Step 5: Outer: ∫02πdθ=2π\int_0^{2\pi} d\theta = 2\pi
Step 6: Product: 13⋅2⋅2π=4π3\frac{1}{3} \cdot 2 \cdot 2\pi = \frac{4\pi}{3}
Answer: 4π3\dfrac{4\pi}{3}

Step 1: Switch to cylindrical: 0≤r≤10 \leq r \leq 1, 0≤θ≤2π0 \leq \theta \leq 2\pi, 0≤z≤20 \leq z \leq 2
Step 2: Integral = ∫02π∫01∫02z⋅r dz dr dθ\int_0^{2\pi} \int_0^1 \int_0^2 z \cdot r \, dz \, dr \, d\theta
Step 3: Inner: ∫02z dz=2\int_0^2 z \, dz = 2
Step 4: Middle: ∫012r dr=1\int_0^1 2r \, dr = 1
Step 5: Outer: ∫02π1 dθ=2π\int_0^{2\pi} 1 \, d\theta = 2\pi
Answer: 2π2\pi

Frequently Asked Questions

Use cylindrical when the region has rotational symmetry around the z-axis but no special radial structure (cylinders, paraboloids, cones above/below a disk). Use spherical when the region is bounded by spheres, cones from the origin, or has full 3D radial symmetry (balls, spherical shells).

The Jacobian is the determinant that adjusts the volume element when changing coordinates. In cylindrical it equals r, in spherical it equals ρ² sin φ. Without it, the integral measures the wrong volume.

Look at the region: integrate the variable with bounds depending on others (innermost) first, then move outward. The outermost variable must have constant bounds. If one order leads to ugly bounds, swap the order using a sketch of the region.

Yes, if the integrand can be negative. For volume calculations the integrand is 1 and the answer is always positive. For physical quantities like signed flux or net force, negative values are possible and meaningful.

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