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.

Why the choice of coordinates matters: the limits of a triple integral describe the solid, and a solid whose boundary is a coordinate surface (ρ=R\rho = R, r=2r = 2, z=4−r2z = 4 - r^2) gets constant or simple limits. The volume element changes because the little coordinate box is no longer a cube: in cylindrical coordinates its θ\theta-edge is an arc of length r dθr\,d\theta. In spherical coordinates the θ\theta-edge sits at distance ρsin⁡ϕ\rho\sin\phi from the axis and has length ρsin⁡ϕ dθ\rho\sin\phi\,d\theta, while the ϕ\phi-edge has length ρ dϕ\rho\,d\phi; multiplying by dρd\rho gives ρ2sin⁡ϕ\rho^2\sin\phi (proved below).

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.

Why the three integrals separate: every limit is a constant and the integrand ρ2sin⁡ϕ\rho^2\sin\phi is a product of a function of ρ\rho and a function of ϕ\phi. So the triple integral is (∫02πdθ)(∫0πsin⁡ϕ dϕ)(∫0Rρ2 dρ)=2π⋅2⋅R33\left(\int_0^{2\pi}d\theta\right)\left(\int_0^\pi\sin\phi\,d\phi\right)\left(\int_0^R\rho^2\,d\rho\right) = 2\pi\cdot2\cdot\tfrac{R^3}{3}. The middle factor 2=[−cos⁡ϕ]0π2 = [-\cos\phi]_0^\pi is the fraction of the sphere's surface swept out by ϕ\phi, counted in the right units: ∫0πsin⁡ϕ dϕ∫02πdθ=4π\int_0^\pi\sin\phi\,d\phi\int_0^{2\pi}d\theta = 4\pi is the area of the unit sphere.

Example
A paraboloid dome
424 − r²
The dome z=4−r2z = 4 - r^2 over the disk r≤2r \le 2; the dashed column at r=1r = 1 has height 4−r2=34 - r^2 = 3 (heights drawn at 0.60.6 scale).

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.

Where the limits come from: the dome meets the floor z=0z = 0 where 4−r2=04 - r^2 = 0, so its shadow is the disk r≤2r \le 2. Above the point (r,θ)(r, \theta) the solid runs from z=0z = 0 up to z=4−r2z = 4 - r^2, which gives the factor 4−r24 - r^2; the volume element supplies one more rr. Then ∫02(4r−r3) dr=8−4=4\int_0^2 (4r - r^3)\,dr = 8 - 4 = 4, and the θ\theta-integral multiplies by 2π2\pi.

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
z = 1z = r
The solid between the cone z=rz = r and the plane z=1z = 1.

Find the volume of the solid above the cone z=rz = r and below the plane z=1z = 1.

Show the answer

The cone meets the plane where r=1r = 1, so the shadow is the unit disk. Above a point at radius rr the solid runs from z=rz = r up to z=1z = 1, a height of 1−r1 - r. With the volume element, ∫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\pi3. That is one third of the enclosing cylinder of radius 11 and height 11, whose volume is π\pi: the cone rule, derived rather than remembered.

Setting Up Triple Integrals in Cartesian Coordinates

Tip
Limits from the inside out

Write ∭Ef dV=∫ab ⁣∫g1(x)g2(x) ⁣∫h1(x,y)h2(x,y)f dz dy dx\displaystyle\iiint_E f\,dV = \int_a^b\!\int_{g_1(x)}^{g_2(x)}\!\int_{h_1(x, y)}^{h_2(x, y)} f\,dz\,dy\,dx. The innermost limits are the bottom and top surfaces of EE above the point (x,y)(x, y); the middle limits describe the shadow of EE on the xyxy-plane; the outer limits are the extent of that shadow in xx. Each limit may depend only on the variables integrated outside it.

Why this works: fix (x,y)(x, y), and the vertical line through it meets EE in a segment running from the bottom surface to the top one; the innermost integral adds up ff along that segment. The points (x,y)(x, y) whose vertical line meets EE at all form the shadow, a plane region, and from there on the problem is an ordinary double integral.

Example
Example: a tetrahedron
xyz244
The tetrahedron under 2x+y+z=42x + y + z = 4, its shadow (dashed) and one column, at (x,y)=(12,1)(x, y) = \left(\tfrac12, 1\right), of height 4−1−1=24 - 1 - 1 = 2.

Find the volume of the solid bounded by the coordinate planes and 2x+y+z=42x + y + z = 4. The top surface is z=4−2x−yz = 4 - 2x - y. Setting z=0z = 0 gives the shadow, the triangle 2x+y≤42x + y \le 4 in the first quadrant: yy from 00 to 4−2x4 - 2x, and xx from 00 to 22. So V=∫02 ⁣∫04−2x ⁣∫04−2x−ydz dy dxV = \displaystyle\int_0^2\!\int_0^{4 - 2x}\!\int_0^{4 - 2x - y} dz\,dy\,dx.

The inner integral is the column height 4−2x−y4 - 2x - y. The middle one is ∫04−2x(4−2x−y) dy=12(4−2x)2\int_0^{4 - 2x}(4 - 2x - y)\,dy = \tfrac12(4 - 2x)^2, the area of the cross-section at xx. The outer one is 12∫02(4−2x)2 dx=12⋅323=163\tfrac12\int_0^2 (4 - 2x)^2\,dx = \tfrac12\cdot\tfrac{32}{3} = \tfrac{16}{3}. Check: a tetrahedron with edges 2,4,42, 4, 4 along the axes has volume 16⋅2⋅4⋅4=163\tfrac16\cdot2\cdot4\cdot4 = \tfrac{16}{3}. With the integrand xx, the same limits give ∫02x⋅12(4−2x)2 dx=83\int_0^2 x\cdot\tfrac12(4 - 2x)^2\,dx = \tfrac83, so the average of xx over the solid is 8/316/3=12\tfrac{8/3}{16/3} = \tfrac12: a quarter of the way along the xx-edge, as for every tetrahedron's centroid.

Example
Example: a curved shadow

Find the volume of the solid 0≤z≤y0 \le z \le y over the region x2≤y≤1x^2 \le y \le 1. The bottom is z=0z = 0 and the top is z=yz = y; the shadow lies between the parabola y=x2y = x^2 and the line y=1y = 1, so −1≤x≤1-1 \le x \le 1. Then V=∫−11 ⁣∫x21 ⁣∫0ydz dy dxV = \displaystyle\int_{-1}^{1}\!\int_{x^2}^{1}\!\int_0^{y} dz\,dy\,dx.

The inner integral gives yy, the middle ∫x21y dy=1−x42\int_{x^2}^1 y\,dy = \tfrac{1 - x^4}{2}, and the outer ∫−111−x42 dx=1−15=45\int_{-1}^1 \tfrac{1 - x^4}{2}\,dx = 1 - \tfrac15 = \tfrac45. The same solid in the order dx dz dydx\,dz\,dy: for fixed yy and zz, xx runs from −y-\sqrt y to y\sqrt y, while 0≤z≤y≤10 \le z \le y \le 1, so V=∫01 ⁣∫0y2y dz dy=∫012y3/2 dy=45V = \int_0^1\!\int_0^y 2\sqrt y\,dz\,dy = \int_0^1 2y^{3/2}\,dy = \tfrac45. Any of the six orders works; the best one makes the limits simplest.

Side note
Where it goes wrong

Limits that ignore the nesting. For the tetrahedron, letting yy run from 00 to 44 for every xx (instead of up to 4−2x4 - 2x) gives ∫02 ⁣∫04(4−2x−y) dy dx=∫02(8−8x) dx=0\int_0^2\!\int_0^4 (4 - 2x - y)\,dy\,dx = \int_0^2 (8 - 8x)\,dx = 0. The integral added the negative "heights" 4−2x−y4 - 2x - y beyond the slanted face, and they cancelled the real volume exactly. A middle limit that should depend on xx but does not is the most common setup error.

Try it
Try it: under a plane

Find the volume of the solid in the first octant under the plane x+2y+3z=6x + 2y + 3z = 6 by setting up and evaluating a triple integral.

Show the answer

The top surface is z=6−x−2y3z = \tfrac{6 - x - 2y}{3}. Its shadow, from z=0z = 0, is the triangle x+2y≤6x + 2y \le 6 with x,y≥0x, y \ge 0: yy from 00 to 6−x2\tfrac{6 - x}{2} and xx from 00 to 66. The middle integral is ∫0(6−x)/26−x−2y3 dy=(6−x)212\int_0^{(6 - x)/2}\tfrac{6 - x - 2y}{3}\,dy = \tfrac{(6 - x)^2}{12}, and the outer one is ∫06(6−x)212 dx=7212=6\int_0^6 \tfrac{(6 - x)^2}{12}\,dx = \tfrac{72}{12} = 6. Check with the tetrahedron formula: edges 66, 33, 22 along the axes give 16⋅6⋅3⋅2=6\tfrac16\cdot6\cdot3\cdot2 = 6.

Cones, Shells, and Spherical Limits

Tip
Reading spherical limits

In spherical coordinates, ρ\rho is the distance from the origin, ϕ\phi the angle down from the positive zz-axis (0≤ϕ≤π0 \le \phi \le \pi) and θ\theta the longitude. A ball is ρ≤R\rho \le R, a shell is a≤ρ≤ba \le \rho \le b, the cone z=x2+y2z = \sqrt{x^2 + y^2} is ϕ=π4\phi = \tfrac\pi4, and "above the cone" means ϕ≤π4\phi \le \tfrac\pi4. The volume element is dV=ρ2sin⁡ϕ dρ dϕ dθdV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta.

Why cones are easy here: on the cone z=x2+y2z = \sqrt{x^2 + y^2}, z=ρcos⁡ϕz = \rho\cos\phi and x2+y2=ρsin⁡ϕ\sqrt{x^2 + y^2} = \rho\sin\phi, so cos⁡ϕ=sin⁡ϕ\cos\phi = \sin\phi and ϕ=π4\phi = \tfrac\pi4 at every point. The cone is a single coordinate surface, just as a sphere is, and a region bounded by spheres and cones becomes a box in (ρ,ϕ,θ)(\rho, \phi, \theta).

Example
Example: an ice-cream cone
rzφφ = π/4ρ = 2
A slice through the zz-axis: the solid lies inside ρ=2\rho = 2 and within the angle ϕ≤π4\phi \le \tfrac\pi4 of the axis.

Find the volume inside the sphere ρ=2\rho = 2 and above the cone ϕ=π4\phi = \tfrac\pi4. The limits are constants: ∫02π ⁣∫0π/4 ⁣∫02ρ2sin⁡ϕ dρ dϕ dθ=2π(1−cos⁡π4)⋅83\displaystyle\int_0^{2\pi}\!\int_0^{\pi/4}\!\int_0^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\left(1 - \cos\tfrac\pi4\right)\cdot\tfrac83.

So V=2π(1−22)83=8π3(2−2)≈4.91V = 2\pi\left(1 - \tfrac{\sqrt2}{2}\right)\tfrac83 = \tfrac{8\pi}{3}\left(2 - \sqrt2\right) \approx 4.91. Check the proportion: the whole ball has volume 32π3≈33.5\tfrac{32\pi}{3} \approx 33.5, and the cone takes the fraction 1−cos⁡(π/4)2≈0.146\tfrac{1 - \cos(\pi/4)}{2} \approx 0.146 of it, because ∫0π/4sin⁡ϕ dϕ\int_0^{\pi/4}\sin\phi\,d\phi is that fraction of ∫0πsin⁡ϕ dϕ=2\int_0^\pi\sin\phi\,d\phi = 2.

Example
Example: a shell with varying density

A spherical shell 1≤ρ≤21 \le \rho \le 2 has density 1ρ\tfrac1\rho. Its mass is ∫02π ⁣∫0π ⁣∫121ρ ρ2sin⁡ϕ dρ dϕ dθ=2π⋅2⋅∫12ρ dρ\displaystyle\int_0^{2\pi}\!\int_0^\pi\!\int_1^2 \tfrac1\rho\,\rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot2\cdot\int_1^2\rho\,d\rho.

That is 4π⋅32=6π4\pi\cdot\tfrac32 = 6\pi. Check by layers: a thin shell at radius ρ\rho has area 4πρ24\pi\rho^2 and density 1ρ\tfrac1\rho, so it carries mass 4πρ4\pi\rho per unit of thickness, and ∫124πρ dρ=6π\int_1^2 4\pi\rho\,d\rho = 6\pi. The spherical integral is exactly this layer-by-layer sum.

Example
Example: a moment of inertia in cylindrical coordinates

The moment of inertia of the solid cylinder r≤1r \le 1, 0≤z≤20 \le z \le 2 (density 11) about its axis is ∭(x2+y2) dV=∫02π ⁣∫01 ⁣∫02r2⋅r dz dr dθ\iiint (x^2 + y^2)\,dV = \displaystyle\int_0^{2\pi}\!\int_0^1\!\int_0^2 r^2\cdot r\,dz\,dr\,d\theta.

That is 2π⋅14⋅2=π2\pi\cdot\tfrac14\cdot2 = \pi. The mass is the volume, 2π2\pi, so the answer matches the engineering formula I=12MR2=12⋅2π⋅1=πI = \tfrac12MR^2 = \tfrac12\cdot2\pi\cdot1 = \pi. The integrand x2+y2=r2x^2 + y^2 = r^2 is the squared distance to the axis, which is why cylindrical coordinates fit.

Side note
Where it goes wrong

Mixing up ϕ\phi and θ\theta. If ϕ\phi runs from 00 to 2π2\pi, the "volume" of a ball, ∫02π ⁣∫02π ⁣∫0Rρ2sin⁡ϕ dρ dϕ dθ\int_0^{2\pi}\!\int_0^{2\pi}\!\int_0^R\rho^2\sin\phi\,d\rho\,d\phi\,d\theta, comes out 00, because ∫02πsin⁡ϕ dϕ=0\int_0^{2\pi}\sin\phi\,d\phi = 0. The angle ϕ\phi is measured down from the north pole and stops at the south pole, ϕ=π\phi = \pi; only θ\theta goes all the way round. The companion slip is writing sin⁡θ\sin\theta instead of sin⁡ϕ\sin\phi in dVdV.

Try it
Try it: half a shell

Find the volume of the region between the spheres ρ=1\rho = 1 and ρ=2\rho = 2 that lies above the xyxy-plane.

Show the answer

Above the xyxy-plane means 0≤ϕ≤π20 \le \phi \le \tfrac\pi2; θ\theta goes all the way round and 1≤ρ≤21 \le \rho \le 2. So V=∫02π ⁣∫0π/2 ⁣∫12ρ2sin⁡ϕ dρ dϕ dθ=2π⋅1⋅8−13=14π3V = \int_0^{2\pi}\!\int_0^{\pi/2}\!\int_1^2 \rho^2\sin\phi\,d\rho\,d\phi\,d\theta = 2\pi\cdot1\cdot\tfrac{8 - 1}{3} = \tfrac{14\pi}{3}. Check: the full shell is 43π(8−1)=28π3\tfrac43\pi(8 - 1) = \tfrac{28\pi}{3}, and half of it is 14π3\tfrac{14\pi}{3}.

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} ✓.