Engineering Math Practice

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.

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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.

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