Questions: Triple Integrals in Spherical Coordinates

Question 1

A student evaluates ∫∫∫_B f(ρ) dV over a solid ball B by writing ∫₀²π ∫₀π ∫₀^R f(ρ) dρ dφ dθ. What error has she made?

• A. She should use cylindrical coordinates, not spherical, for a ball
• B. She forgot the Jacobian factor ρ² sin φ — the correct volume element is dV = ρ² sin φ dρ dφ dθ, not dρ dφ dθ
• C. The limits for φ should range from 0 to 2π, not 0 to π
• D. The outer integral over θ should come last, not first

Answer: (B)

When changing variables in a multiple integral, you must multiply by the Jacobian determinant of the transformation. For spherical coordinates, this is ρ² sin φ. Omitting it is equivalent to pretending the coordinate transformation causes no distortion, which is false — near the poles, a given dθ sweeps out very little volume, while near the equator it sweeps out much more. The factor ρ² sin φ correctly accounts for this varying distortion. Omitting it gives a numerically wrong answer.

Question 2

Why does the factor sin φ appear in the spherical volume element dV = ρ² sin φ dρ dφ dθ?

• A. It is a correction for the curvature of the integrand near the origin
• B. Near the poles (φ ≈ 0 or π), a small dθ sweeps out very little arc length, while near the equator (φ = π/2) the same dθ sweeps out the full ρ dθ — sin φ scales the azimuthal contribution correctly with latitude
• C. It converts between radians and degrees in the angular variables
• D. It ensures the integral over a full sphere gives the right surface area instead of volume

Answer: (B)

The sin φ factor reflects the geometry of latitude circles. At the poles, longitude lines converge — a step dθ in azimuthal angle covers negligible distance. At the equator, the same dθ covers maximum distance. The arc length in the θ-direction at polar angle φ and radius ρ is ρ sin φ dθ, giving the full volume element ρ dρ · ρ dφ · ρ sin φ dθ = ρ² sin φ dρ dφ dθ. This is the geometric explanation of why sin φ must appear.

Question 3

In spherical coordinates, the volume element is simply dV = dρ dφ dθ, analogous to dV = dx dy dz in Cartesian coordinates.

• T. True
• F. False

Answer: False

Unlike Cartesian coordinates, where the coordinate grid is orthonormal and undistorted, spherical coordinates have a grid that stretches and compresses depending on position. Changing variables in any integral requires multiplying by the Jacobian of the transformation to account for this distortion. For spherical coordinates, the Jacobian is ρ² sin φ, giving dV = ρ² sin φ dρ dφ dθ. Forgetting this factor is the most common error in spherical coordinate integration and produces completely wrong numerical results.

Question 4

A function of the form f(x,y,z) = g(x²+y²+z²) becomes much simpler in spherical coordinates, because x²+y²+z² = ρ², reducing f to a function of ρ alone.

• T. True
• F. False

Answer: True

This is precisely the type of function that makes spherical coordinates advantageous. When f depends only on distance from the origin, it becomes f(ρ) in spherical coordinates. The angular integrals ∫₀²π dθ and ∫₀π sin φ dφ then factor out completely, evaluating to 2π and 2 respectively (product = 4π), and the entire 3D integral reduces to 4π ∫₀^R f(ρ) ρ² dρ. Functions like e^(−(x²+y²+z²)) = e^(−ρ²) that are extremely hard in Cartesian become one-dimensional in spherical.

Question 5

Explain why converting a triple integral over a ball-shaped region to spherical coordinates can reduce what looks like an intractable Cartesian problem to a simple calculation. What specific features of spherical coordinates enable this?

Model answer: Spherical coordinates match the ball's geometry: the region 0 ≤ ρ ≤ R, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π describes a ball as a simple rectangular box in (ρ, φ, θ) space, avoiding the complicated square-root limits needed in Cartesian coordinates. Functions that depend on x²+y²+z² (distance from origin) become functions of ρ alone. The angular integrals are constants that factor out. In contrast, integrating over a ball in Cartesian requires limits like −√(R²−x²−y²) to √(R²−x²−y²) for z, coupling all three variables and making the integrand much harder to evaluate.

The key principle is that coordinate systems should match the symmetry of the region and integrand. Whenever a region has spherical symmetry (defined by r = constant) or a function depends only on distance from a point, spherical coordinates will simplify the integral dramatically — at the cost of requiring the Jacobian factor ρ² sin φ, which must never be forgotten.