Engineering Math PracticeGet the pack

Triple Integrals in Spherical Coordinates

Convert a triple integral over a ball to spherical coordinates and evaluate it.

The idea. Spherical coordinates locate a point by a distance and two angles. This is the payoff coordinate system for anything symmetric around a point — spheres, cones, ice-cream-cone solids. Three conversions and the Jacobian that defines this entire section:

x=ρsin⁡ϕcos⁡θ,y=ρsin⁡ϕsin⁡θ,z=ρcos⁡ϕ,x2+y2+z2=ρ2x = \rho\sin\phi\cos\theta, \qquad y = \rho\sin\phi\sin\theta, \qquad z = \rho\cos\phi, \qquad x^2+y^2+z^2 = \rho^2

Spherical substitutions

  • ρ\rhodistance from the origin; ρ ≥ 0.
  • ϕ\phiangle down from the positive z-axis; 0 ≤ φ ≤ π (latitude from the north pole).
  • θ\thetaangle around the z-axis in the xy-plane; 0 ≤ θ ≤ 2π (longitude).
  • x2+y2+z2=ρ2x^2+y^2+z^2=\rho^2collapses spherical integrands.

Ready to put it into practice?

Start practicing

The volume element:

dV=ρ2sin⁡ϕ  dρ dϕ dθdV = \rho^2\sin\phi\; d\rho\, d\phi\, d\theta

Volume element in spherical coordinates

  • dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi\, d\rho\, d\phi\, d\thetathe volume element; ρ²sin φ is the Jacobian, never optional.

That ρ² sin φ is the whole difficulty of the section. It is not optional, it is not just ρ², and it is not just sin φ — it's both, multiplied. Forgetting it, or dropping half of it, is the single most common spherical mistake, more so even than the missing r in polar. Burn it in: rho-squared sine phi.

What the three coordinates mean — and this is where people get turned around. ρ (rho) is distance from the origin, always ≥ 0. φ (phi) is the angle down from the positive z-axis, ranging 0 to π: φ = 0 points straight up, φ = π/2 is the xy-plane, φ = π points straight down — think latitude, measured from the north pole. θ (theta) is the same angle as always, around the z-axis in the xy-plane, ranging 0 to 2π — think longitude. The trap beyond the Jacobian: φ and θ are easy to swap. φ is the up-down tilt (from the top), θ is the around-spin. Getting them backwards wrecks the bounds.

The tell — reach for spherical when: the solid is a sphere or a piece of one, there's symmetry about the origin, or the integrand contains x² + y² + z² (collapses to ρ²). A full sphere is the cleanest case: ρ ∈ [0, R], φ ∈ [0, π], θ ∈ [0, 2π].

The general form, with f rewritten in spherical coordinates:

∭Ef dV=∫αβ∫ϕ1ϕ2∫ρ1ρ2f ρ2sin⁡ϕ  dρ dϕ dθ\iiint_E f\, dV = \int_\alpha^\beta \int_{\phi_1}^{\phi_2} \int_{\rho_1}^{\rho_2} f\,\rho^2 \sin\phi\; d\rho\, d\phi\, d\theta

Triple integral in spherical coordinates

Worked example

∭E(x2+y2+z2) dV,E:x2+y2+z2≤4\iiint_E (x^2+y^2+z^2)\, dV, \qquad E: x^2+y^2+z^2 \le 4
  1. x2+y2+z2≤4  ⟹  0≤ρ≤2,  0≤ϕ≤π,  0≤θ≤2πx^2+y^2+z^2 \le 4 \;\Longrightarrow\; 0 \le \rho \le 2,\; 0 \le \phi \le \pi,\; 0 \le \theta \le 2\pi

    Convert the region. A ball of radius 2 centered at the origin: ρ from 0 to 2, φ from 0 to π, θ from 0 to 2π. Convert the integrand: x² + y² + z² = ρ².

  2. ∫02π∫0π∫02ρ2⋅ρ2sin⁡ϕ  dρ dϕ dθ=∫02π∫0π∫02ρ4sin⁡ϕ  dρ dϕ dθ\int_0^{2\pi} \int_0^\pi \int_0^2 \rho^2 \cdot \rho^2 \sin\phi\; d\rho\, d\phi\, d\theta = \int_0^{2\pi} \int_0^\pi \int_0^2 \rho^4\sin\phi\; d\rho\, d\phi\, d\theta

    Set up — attach the full Jacobian. The integrand ρ² times the Jacobian ρ²sin φ gives ρ⁴sin φ. That multiplication is the whole game.

  3. ∫02ρ4 dρ=ρ55∣02=325\int_0^2 \rho^4\, d\rho = \frac{\rho^5}{5}\bigg|_0^2 = \frac{32}{5}

    Innermost, over ρ.

  4. ∫0πsin⁡ϕ dϕ=[−cos⁡ϕ]0π=−(−1)−(−1)=2\int_0^\pi \sin\phi\, d\phi = [-\cos\phi]_0^\pi = -(-1) - (-1) = 2

    Middle, over φ.

  5. ∫02π1 dθ=2π\int_0^{2\pi} 1\, d\theta = 2\pi

    Outermost, over θ.

  6. 325⋅2⋅2π=128π5\frac{32}{5} \cdot 2 \cdot 2\pi = \frac{128\pi}{5}

    Multiply the three. Because the bounds are all constants and the integrand separated, this became three independent single integrals multiplied together — the cleanest kind of spherical problem, and a good template.

Answer:

128π5\dfrac{128\pi}{5}

How to solve one of these

  1. 1.

    Spot the sphere. Spherical region, symmetry about the origin, or x2+y2+z2x^2+y^2+z^2 in the integrand. That's the signal.

  2. 2.

    Set the bounds. Full sphere: ρ∈[0,R]\rho \in [0,R], ϕ∈[0,π]\phi \in [0,\pi], θ∈[0,2π]\theta \in [0,2\pi]. A hemisphere or cone narrows ϕ\phi; a wedge narrows θ\theta.

  3. 3.

    Convert the integrand. Use x2+y2+z2=ρ2x^2+y^2+z^2 = \rho^2 where possible; otherwise substitute each of x,y,zx, y, z in full.

  4. 4.

    Attach the Jacobian. Write ρ2sin⁡ϕ dρ dϕ dθ\rho^2 \sin\phi\, d\rho\, d\phi\, d\theta. Multiply it into the integrand now, before integrating. This is the step to never skip.

  5. 5.

    Integrate inside-out. ρ\rho, then ϕ\phi, then θ\theta. When all bounds are constant and the integrand separates, do the three as independent integrals and multiply.

  6. 6.

    Sanity-check. The volume of a ball should come out 43πR3\frac{4}{3}\pi R^3. If a volume calculation misses that, you probably dropped part of the Jacobian.

dV=ρ2sin⁡ϕ  dρ dϕ dθdV = \rho^2\sin\phi\; d\rho\, d\phi\, d\theta

The volume element in spherical coordinates: ρ²sin φ is the Jacobian, and it's never optional — not just ρ², not just sin φ, but both, multiplied.

Ready to put it into practice?

Start practicing

Want the rest of Calc III? Leave your email and I'll tell you the day the pack's live.