Difficulty: Intermediate | Prerequisites: MATH 215 through §15.6 (double integrals, polar coordinates)
Cylindrical coordinates (r, θ, z) extend polar coordinates into 3D and work best for problems with circular symmetry about the z-axis. Spherical coordinates (ρ, θ, φ) describe points by distance from the origin, azimuthal angle, and polar angle, and they simplify integration over balls, cones, and spheres. Both systems require a Jacobian factor in the volume element when setting up triple integrals.
Cylindrical coordinates (r, θ, z)
A coordinate system where r is the distance from the z-axis, θ is the angle measured from the positive x-axis in the xy-plane, and z is the height. Think of it as polar coordinates for the horizontal plane, with z tacked on for height.
Spherical coordinates (ρ, θ, φ)
A system where ρ is the distance from the origin, θ is the same azimuthal angle as in cylindrical coordinates, and φ is the angle measured down from the positive z-axis. In simple terms, ρ tells you how far from the origin, φ tells you how far from the north pole, and θ tells you the compass direction.
Jacobian (for coordinate transformations)
The determinant of the matrix of partial derivatives that converts a volume (or area) element from one coordinate system to another. In simple terms, it is the correction factor that accounts for how the coordinate grid stretches or compresses space.
Volume element in cylindrical coordinates: dV = r dr dθ dz
The factor r appears because annular slices at larger radius sweep out more volume. Think of it as the same r that appears in polar area elements, just extended into 3D.
Volume element in spherical coordinates: dV = ρ² sin φ dρ dθ dφ
The ρ² sin φ factor accounts for the way spherical shells expand with radius (ρ²) and the way latitude bands shrink near the poles (sin φ). This is the single most important formula to memorise for spherical integration.
Density function
A function δ(x, y, z) giving mass per unit volume at each point. When a problem says "density is proportional to the distance from the origin," it means δ = kρ for some constant k determined by given conditions.
Setting one coordinate equal to another (or to a constant) produces recognisable 3D surfaces. Practising this builds the geometric intuition needed to choose the right coordinate system for a given region.
r = θ (cylindrical): At each height z, the cross-section is an Archimedean spiral. The full surface is this spiral extruded vertically, forming a helical sheet.
ρ = θ (spherical): The distance from the origin equals the azimuthal angle. This produces a spiral-like surface in 3D that winds outward as θ increases, but with a spherical geometry rather than a flat one.
r = ρ (cylindrical r equals spherical ρ): Since ρ² = r² + z², the equation r = ρ implies r² = r² + z², so z = 0. This is the xy-plane (away from the origin). The condition also holds trivially at the origin.
θ = φ (spherical): The azimuthal angle equals the polar angle. This defines a surface that is a cone-like shape whose opening angle changes with direction, tracing out a curve on each sphere of constant ρ.
The ball of radius 1 centred at (0, 0, 1) is described in Cartesian coordinates by x² + y² + z² − 2z ≤ 0 (complete the square: x² + y² + (z − 1)² ≤ 1).
To convert this to spherical coordinates, substitute x² + y² + z² = ρ² and z = ρ cos φ:
The inequality becomes ρ² − 2ρ cos φ ≤ 0, which factors as ρ(ρ − 2 cos φ) ≤ 0.
Since ρ ≥ 0, this gives 0 ≤ ρ ≤ 2 cos φ.
For the ball to exist (ρ > 0), we need cos φ > 0, so 0 ≤ φ < π/2.
θ ranges over its full period: 0 ≤ θ < 2π.
The key insight: when a sphere is not centred at the origin, the upper bound on ρ becomes a function of the angles.
Given a density proportional to distance from the origin with maximum density 2, first determine the proportionality constant. The maximum distance from the origin within the ball centred at (0, 0, 1) is 2 (the point (0, 0, 2)), so δ = 2 · (ρ/2) = ρ.
Mass is then:
M = ∫∫∫_E ρ · ρ² sin φ dρ dθ dφ = ∫₀²π ∫₀^(π/2) ∫₀^(2cos φ) ρ³ sin φ dρ dφ dθ
Centre of mass: by symmetry about the z-axis, x̄ = 0 and ȳ = 0. Only z̄ requires computation, using z = ρ cos φ in the integrand.
Why the ball centred at (0, 1, 0) is harder: that ball's Cartesian equation involves y, which in spherical coordinates is ρ sin φ sin θ. The resulting bound on ρ depends on both φ and θ, making the integral substantially more complex. The (0, 0, 1) ball is simpler because z = ρ cos φ involves only φ.
Start with a sphere of radius R. Bore a cylindrical hole along the z-axis, leaving a band of height h.
The cylinder's radius a satisfies a² + (h/2)² = R² (from the geometry where the cylinder meets the sphere), so a² = R² − h²/4.
Use cylindrical coordinates. The sphere is r² + z² = R², and the remaining region has a ≤ r ≤ √(R² − z²), with −h/2 ≤ z ≤ h/2.
The resulting volume is V = πh³/6. This depends only on h, not on R. A tall, thin ring cut from a small sphere has the same volume as a short, wide ring cut from a large sphere, so long as the band heights match.
A wedge cut from a sphere of radius R by two planes intersecting along a diameter at angle π/5, with density proportional to distance from the origin (maximum density 2):
The maximum distance from the origin is R (the sphere's surface), so δ = (2/R)ρ.
In spherical coordinates, the wedge spans: 0 ≤ ρ ≤ R, 0 ≤ θ ≤ π/5, 0 ≤ φ ≤ π.
Mass = ∫₀^(π/5) ∫₀^π ∫₀^R (2ρ/R) · ρ² sin φ dρ dφ dθ = (π/5) · 2 · (2/R) · R⁴/4 = πR³/5.
x = r cos θ, y = r sin θ, z = z
r² = x² + y², tan θ = y/x
x = ρ sin φ cos θ, y = ρ sin φ sin θ, z = ρ cos φ
ρ² = x² + y² + z²
r = ρ sin φ, z = ρ cos φ, θ = θ
ρ² = r² + z²
Cylindrical: dV = r dr dθ dz
Spherical: dV = ρ² sin φ dρ dθ dφ
Full cylinder of radius a, height h: 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π, 0 ≤ z ≤ h
Full sphere of radius R: 0 ≤ ρ ≤ R, 0 ≤ θ ≤ 2π, 0 ≤ φ ≤ π
Upper hemisphere: same but 0 ≤ φ ≤ π/2
Sphere centred at (0, 0, a): 0 ≤ ρ ≤ 2a cos φ, 0 ≤ φ ≤ π/2, 0 ≤ θ ≤ 2π
Forgetting the Jacobian factor. Students write dV = dρ dθ dφ in spherical coordinates, omitting the ρ² sin φ. Every coordinate change has a Jacobian; in spherical it is ρ² sin φ, in cylindrical it is r.
Confusing φ and θ. In the physics convention, the meanings of θ and φ are swapped relative to the maths convention used in Stewart and in MATH 215. Always check which convention your course uses. In this course, φ is the angle from the positive z-axis (0 ≤ φ ≤ π) and θ is the azimuthal angle in the xy-plane.
Using φ from 0 to 2π. The polar angle φ runs from 0 to π (north pole to south pole). The full 2π range belongs to θ. Setting φ up to 2π double-counts the sphere.
Assuming ρ always runs from 0 to R. When a sphere is not centred at the origin, the upper limit of ρ depends on the angles. For the ball centred at (0, 0, 1), the bound is ρ ≤ 2 cos φ, not ρ ≤ 1.
⚠️ Setting up the correct bounds in spherical coordinates for off-centre spheres is a classic exam problem. Practise converting x² + y² + z² − 2z ≤ 0 to spherical; the technique generalises to any sphere centred on the z-axis.
⚠️ The napkin ring result (V = πh³/6 depending only on h) is a well-known "surprising" result that examiners enjoy asking about. Know the derivation, not just the formula.
⚠️ Mass and centre of mass problems require setting up density × volume element correctly. If density is proportional to distance from the origin, write δ = kρ and determine k from the given maximum.
⚠️ Recognising when cylindrical coordinates are better than spherical (and vice versa) is a core exam skill. Cylinders, vertical symmetry, and flat tops/bottoms favour cylindrical. Spheres, cones, and radial density favour spherical.
True or false: The volume element in spherical coordinates is dV = ρ sin φ dρ dθ dφ. (False: it is ρ² sin φ dρ dθ dφ.)
Fill in the blank: For a full sphere, φ ranges from ___ to ___. (0 to π.)
True or false: A ball of radius 1 centred at (0, 0, 1), when written in spherical coordinates, has a constant upper bound on ρ. (False: the bound is ρ ≤ 2 cos φ, which depends on φ.)
Fill in the blank: The volume of a napkin ring of band height h bored from a sphere is ___. (πh³/6.)
True or false: In the maths convention used in MATH 215, φ is the angle in the xy-plane. (False: φ is the angle from the positive z-axis. θ is the xy-plane angle.)
Q: Write the ball x² + y² + (z − 1)² ≤ 1 in spherical coordinates, specifying all bounds on ρ, θ, and φ.
A: Expand to x² + y² + z² − 2z ≤ 0. Substitute ρ² − 2ρ cos φ ≤ 0, giving 0 ≤ ρ ≤ 2 cos φ, with 0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ 2π.
Q: Set up (but do not evaluate) the integral for the mass of the ball E centred at (0, 0, 1) with radius 1, where the density equals ρ (the spherical distance from the origin).
A: M = ∫₀^(2π) ∫₀^(π/2) ∫₀^(2cos φ) ρ · ρ² sin φ dρ dφ dθ = ∫₀^(2π) ∫₀^(π/2) ∫₀^(2cos φ) ρ³ sin φ dρ dφ dθ.
Q: Explain why the volume of a napkin ring depends only on the band height h and not on the sphere's radius R.
A: The cylinder radius a is determined by R and h through a² = R² − h²/4. When you compute the integral of the remaining volume in cylindrical coordinates, the R terms cancel, leaving V = πh³/6. A larger sphere requires a wider bore to achieve the same band height, and the extra volume removed exactly compensates for the larger starting sphere.
Q: A wedge is cut from a sphere of radius R by two half-planes through a diameter, meeting at angle α. If the density is δ = (2/R)ρ, write the integral for the mass of the wedge.
A: M = ∫₀^α ∫₀^π ∫₀^R (2ρ/R) · ρ² sin φ dρ dφ dθ = (2/R) ∫₀^α ∫₀^π ∫₀^R ρ³ sin φ dρ dφ dθ.
Q: Why is a ball centred at (0, 1, 0) harder to describe in spherical coordinates than one centred at (0, 0, 1)?
A: The ball at (0, 1, 0) has equation x² + (y − 1)² + z² ≤ 1, which expands to ρ² − 2ρ sin φ sin θ ≤ 0. The upper bound ρ ≤ 2 sin φ sin θ depends on both angles, making the integration limits more complex. The (0, 0, 1) ball gives a bound depending only on φ.
This material builds directly on polar coordinates and double integrals from §15.3-15.4. The volume element r dr dθ in polar is the 2D analogue of the cylindrical and spherical volume elements here. The next section (§15.9/16.1) introduces change of variables in multiple integrals, which generalises the idea: cylindrical and spherical coordinate transformations are specific cases of the general Jacobian-based substitution formula.
Centre of mass problems connect to §15.5 (applications of double integrals), extended to three dimensions. The density-times-volume-element pattern is identical; only the coordinate system changes.
cylindrical coordinates, spherical coordinates, triple integrals, volume element, Jacobian, ρ² sin φ, r dr dθ dz, MATH 215, Calculus III, University of Michigan, §15.7, §15.8, multivariable calculus, change of variables, polar coordinates 3D, mass, centre of mass, center of mass, napkin ring theorem, spherical wedge, ball in spherical coordinates, off-centre sphere, density proportional to distance