Triple Integrals & Coordinate Systems

Study Sheet

Triple Integrals & Coordinate Systems

Cylindrical and spherical coordinates

Integrating in Three Dimensions

Tip
Three nests, three coordinate systems

Boxes: integrate dz dy dxdz\,dy\,dx straight through. Solids with circular symmetry: CYLINDRICAL coordinates, dV=r dz dr dθdV = r\,dz\,dr\,d\theta. Balls and cones: SPHERICAL, dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. Choosing coordinates to match the region is most of the work.

Example
The sphere, verified

The ball ρ≤R\rho \le R in spherical: ∫02π ⁣∫0π ⁣∫0Rρ2sin⁡ϕ dρ dϕ dθ=2π⋅2⋅R33=43πR3\int_0^{2\pi}\!\int_0^{\pi}\!\int_0^R \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi \cdot 2 \cdot \tfrac{R^3}{3} = \tfrac43\pi R^3 — the middle-school formula, finally proved.

Example
A paraboloid dome

Under z=4−x2−y2z = 4 - x^2 - y^2: cylindrical gives ∫02π ⁣∫02(4−r2) r dr dθ=8π\int_0^{2\pi}\!\int_0^2 (4 - r^2)\,r\,dr\,d\theta = 8\pi. Forgetting the extra rr gives a wrong answer that LOOKS plausible — always write the volume element first.

Side note
Significance: why three coordinate systems exist

Nature keeps producing round things — planets, pipes, atoms — and Cartesian boxes fit them badly. Cylindrical and spherical coordinates exist because the volume elements rr and ρ2sin⁡ϕ\rho^2\sin\phi, once understood, turn week-long integrals into three lines. The hydrogen atom of quantum mechanics is solved in spherical coordinates for exactly this reason.

Try it
Try it: a cone by the slice

Volume of the cone zz from rr up to 11 over the unit disk (i.e. between z=rz = r and z=1z = 1). Work: ∫02π ⁣∫01(1−r) r dr dθ=2π(12−13)=π3\int_0^{2\pi}\!\int_0^1(1 - r)\,r\,dr\,d\theta = 2\pi\left(\tfrac12 - \tfrac13\right) = \tfrac{\pi}{3} — one third of the enclosing cylinder, the classical cone rule, derived rather than remembered.

Proofs & Why It Matters

Tip
Proof: dV = ρ² sin φ dρ dφ dθ

A small spherical cell has three nearly-perpendicular edges: radial, of length Δρ\Delta\rho; along a meridian, an arc of radius ρ\rho subtending Δϕ\Delta\phi, length ρ Δϕ\rho\,\Delta\phi; and along a parallel, an arc of radius ρsin⁡ϕ\rho\sin\phi (the distance to the zz-axis) subtending Δθ\Delta\theta, length ρsin⁡ϕ Δθ\rho\sin\phi\,\Delta\theta. The cell volume is the product ρ2sin⁡ϕ Δρ Δϕ Δθ\rho^2\sin\phi\,\Delta\rho\,\Delta\phi\,\Delta\theta. ■\blacksquare

Tip
Proof: the sphere volume formula

Integrate the element over the ball of radius RR: ∫02π ⁣ ⁣∫0π ⁣ ⁣∫0Rρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi}\!\!\int_0^{\pi}\!\!\int_0^R \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. The three integrals separate: 2π⋅[−cos⁡ϕ]0π⋅R33=2π⋅2⋅R33=43πR32\pi \cdot \left[-\cos\phi\right]_0^\pi \cdot \tfrac{R^3}{3} = 2\pi\cdot 2\cdot\tfrac{R^3}{3} = \tfrac43\pi R^3. ■\blacksquare

Going Deeper: Explanations & Worked Problems

Concept
Choosing coordinates: read the region, then the integrand

The decision tree: does the region have an axis of circular symmetry? If the boundary involves x2+y2x^2 + y^2 (cylinders, paraboloids, cones over the zz-axis), use CYLINDRICAL — polar in the floor plane with zz kept, dV=r dz dr dθdV = r\,dz\,dr\,d\theta.

If distances from a single point rule (balls, spherical shells, cones from the origin), use SPHERICAL: ρ\rho (distance from origin), ϕ\phi (angle down from the north pole, 00 to π\pi), θ\theta (longitude), with dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta. Then check the integrand cooperates: x2+y2=r2x^2 + y^2 = r^2 in cylindrical; x2+y2+z2=ρ2x^2 + y^2 + z^2 = \rho^2 and z=ρcos⁡ϕz = \rho\cos\phi in spherical. A sphere in Cartesian coordinates costs three nested square-root limits; the SAME sphere in spherical coordinates is a box [0,R]×[0,π]×[0,2π][0,R]\times[0,\pi]\times[0,2\pi] — coordinates are chosen to make the region a box, because boxes separate.

Example
Worked: the paraboloid volume in cylindrical, every step

Volume under z=4−x2−y2z = 4 - x^2 - y^2 above the xyxy-plane.

Step 1 — find the floor region: z=0z = 0 forces x2+y2=4x^2 + y^2 = 4, the disk of radius 22.

Step 2 — write in cylindrical: zz runs from 00 up to 4−r24 - r^2; the volume element contributes the extra rr.

Step 3 — innermost (zz): ∫04−r2dz=4−r2\int_0^{4-r^2}dz = 4 - r^2.

Step 4 — middle (rr): ∫02(4−r2) r dr=∫02(4r−r3)dr=[2r2−r44]02=8−4=4\int_0^2 (4 - r^2)\,r\,dr = \int_0^2 (4r - r^3)dr = \left[2r^2 - \tfrac{r^4}{4}\right]_0^2 = 8 - 4 = 4.

Step 5 — outer (θ\theta): ∫02π4 dθ=8π\int_0^{2\pi}4\,d\theta = 8\pi.

Step 6 — sanity: the enclosing cylinder (radius 22, height 44) has volume 16π16\pi; the paraboloid fills exactly HALF of it — a clean known fact that confirms the arithmetic.

Example
Worked: an off-center integrand in spherical

Compute ∭Bz2 dV\iiint_B z^2\,dV over the ball ρ≤1\rho \le 1.

Step 1 — translate: z=ρcos⁡ϕz = \rho\cos\phi, so z2=ρ2cos⁡2ϕz^2 = \rho^2\cos^2\phi, and with dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta the integrand is ρ4cos⁡2ϕsin⁡ϕ\rho^4\cos^2\phi\sin\phi.

Step 2 — separate the three integrals: (∫01ρ4dρ)(∫0πcos⁡2ϕsin⁡ϕ dϕ)(∫02πdθ)\left(\int_0^1\rho^4d\rho\right)\left(\int_0^\pi\cos^2\phi\sin\phi\,d\phi\right)\left(\int_0^{2\pi}d\theta\right).

Step 3 — evaluate each: ∫01ρ4dρ=15\int_0^1\rho^4d\rho = \tfrac15; for the middle, substitute u=cos⁡ϕu = \cos\phi, du=−sin⁡ϕ dϕdu = -\sin\phi\,d\phi: ∫−11u2du=23\int_{-1}^{1}u^2du = \tfrac23; the last is 2π2\pi.

Step 4 — multiply: 15⋅23⋅2π=4π15\tfrac15\cdot\tfrac23\cdot2\pi = \tfrac{4\pi}{15}.

Step 5 — symmetry check: by symmetry ∭x2=∭y2=∭z2\iiint x^2 = \iiint y^2 = \iiint z^2, and the three must add to ∭ρ2 dV=∫01ρ4dρ⋅4π=4π5\iiint\rho^2\,dV = \int_0^1\rho^4d\rho\cdot4\pi = \tfrac{4\pi}{5}; indeed 3⋅4π15=4π53\cdot\tfrac{4\pi}{15} = \tfrac{4\pi}{5} ✓.