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What is a triple integral?

A triple integral extends the concept of single and double integrals to three dimensions, summing a function f(x,y,z) over a 3D region V. For regions with circular symmetry in the xy-plane, you can switch to cylindrical (polar + z) or spherical coordinates for a simpler setup.

Triple integral formula

The general form is:

∭Vf(x,y,z) dV\iiint_V f(x,y,z)\,dV

In rectangular coordinates:

∫z1z2∫y1(z)y2(z)∫x1(y,z)x2(y,z)f(x,y,z) dx dy dz\int_{z_1}^{z_2}\int_{y_1(z)}^{y_2(z)}\int_{x_1(y,z)}^{x_2(y,z)} f(x,y,z)\,dx\,dy\,dz

In cylindrical coordinates (x=rcos⁡θx=r\cos\theta,y=rsin⁡θy=r\sin\theta, z=zz=z; Jacobian = rr):

∫θ=αβ∫r=r1r2∫z=z1(r,θ)z2(r,θ)f(rcos⁡θ,rsin⁡θ,z) r dz dr dθ\int_{\theta=\alpha}^{\beta}\int_{r=r_1}^{r_2}\int_{z=z_1(r,\theta)}^{z_2(r,\theta)} f(r\cos\theta,r\sin\theta,z)\,r\,dz\,dr\,d\theta

In spherical coordinates (x=ρsin⁡ϕcos⁡θx=\rho\sin\phi\cos\theta,y=ρsin⁡ϕsin⁡θy=\rho\sin\phi\sin\theta,z=ρcos⁡ϕz=\rho\cos\phi; Jacobian = ρ2sin⁡ϕ\rho^2\sin\phi):

∫02π∫0π∫0ρ2(ϕ,θ)f(ρsin⁡ϕcos⁡θ,ρsin⁡ϕsin⁡θ,ρcos⁡ϕ) ρ2sin⁡ϕ dρ dϕ dθ\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\rho_2(\phi,\theta)} f(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi)\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta

How to solve a triple integral step by step

Step 1: Identify f(x,y,z) and the region V; choose rectangular, cylindrical (polar + z), or spherical coordinates.

Step 2: Rewrite dVdV with the correct Jacobian (dx dy dzdx\,dy\,dz,r dr dθ dzr\,dr\,d\theta\,dz orρ2sin⁡ϕ dρ dϕ dθ\rho^2\sin\phi\,d\rho\,d\phi\,d\theta) and set your limits.

Step 3: Integrate the innermost integral first, then proceed outward.

Step 4: Simplify and evaluate to get your final result.

Example: ∭V(x2+y2+z2) dV\iiint_V (x^2 + y^2 + z^2)\,dV over the unit sphere x2+y2+z2≤1x^2 + y^2 + z^2 \le 1

Using spherical coordinates:

x=ρsin⁡ϕcos⁡θy=ρsin⁡ϕsin⁡θz=ρcos⁡ϕdV=ρ2sin⁡ϕ dρ dϕ dθ∭V(x2+y2+z2) dV=∫02π ⁣∫0π ⁣∫01ρ2⋅ρ2sin⁡ϕ dρ dϕ dθ=∫02π ⁣∫0π ⁣∫01ρ4sin⁡ϕ dρ dϕ dθ=4π5 x = \rho\sin\phi\cos\theta \\ y = \rho\sin\phi\sin\theta \\ z = \rho\cos\phi \\ dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta \\[6pt] \iiint_V (x^2 + y^2 + z^2)\,dV = \int_{0}^{2\pi}\!\int_{0}^{\pi}\!\int_{0}^{1} \rho^2 \cdot \rho^2\sin\phi \,d\rho\,d\phi\,d\theta \\ = \int_{0}^{2\pi}\!\int_{0}^{\pi}\!\int_{0}^{1} \rho^4\sin\phi \,d\rho\,d\phi\,d\theta = \frac{4\pi}{5}

Tips & Tricks

  • For cylindrical, remember it’s polar (r, θ) plus z with Jacobian r.
  • Use spherical for perfect spheres or cones—don’t forget the ρ² sin φ factor.
  • Sketch the 3D region and its projection onto the xy-plane or φ–θ plane to get accurate limits.

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