Questions: Applications of Triple Integrals: Volume and Mass

Question 1

You need to compute the volume of a solid ball of radius R centered at the origin. Which coordinate system minimizes computational effort, and why?

• A. Cartesian, because all integrals are straightforward rectangles
• B. Cylindrical, because r² = x² + y² simplifies the radial boundary
• C. Spherical, because the boundary ρ = R is a constant, turning all integration limits into constants
• D. Any coordinate system gives equal effort — the Jacobian compensates exactly

Answer: (C)

In spherical coordinates (ρ, φ, θ), the ball 0 ≤ ρ ≤ R, 0 ≤ φ ≤ π, 0 ≤ θ ≤ 2π has rectangular (constant) limits — the easiest possible case. In Cartesian coordinates, the limits involve nested square roots: z from −√(R²−x²−y²) to √(R²−x²−y²), etc. The Jacobian in spherical coordinates introduces ρ² sin(φ), but the tradeoff is overwhelmingly worthwhile. Matching coordinate system to region symmetry is the central skill of this topic.

Question 2

A solid cylinder has variable density δ(r, θ, z) = r (denser farther from the axis). To compute its moment of inertia about the z-axis, I_z = ∭(x² + y²)δ dV, which coordinate substitution is most natural?

• A. Cartesian, because x and y appear explicitly in the integrand
• B. Spherical, because the cylinder has rotational symmetry
• C. Cylindrical, because x² + y² = r² and the cylinder's boundary is r = constant
• D. No substitution is needed — the moment of inertia formula does not require integration

Answer: (C)

Cylindrical coordinates (r, θ, z) with dV = r dr dθ dz are ideal here for two reasons: (1) x² + y² = r² simplifies the integrand directly, and (2) the cylinder's boundary in cylindrical coordinates is simply r = R (a constant), making the limits rectangular. The density δ = r is also already in terms of r. In Cartesian coordinates, x² + y² stays as-is and the cylindrical boundary becomes x² + y² = R², requiring non-constant limits.

Question 3

The mass of a solid with uniform density δ₀ is equal to δ₀ times its volume, which can be computed as ∭_E dV.

• T. True
• F. False

Answer: True

When density is constant (δ = δ₀), mass = ∭_E δ dV = δ₀ ∭_E dV = δ₀ × Volume. The formula for volume, ∭_E dV (integrating 1 over the region), is correct in any coordinate system — the Jacobian factor in each coordinate system ensures dV is the correct volume element. For variable density, you must include δ(x,y,z) in the integrand.

Question 4

Switching from Cartesian to spherical coordinates for a spherical region changes the conceptual content of the integral (e.g., what 'mass' means), not just the computational form.

• T. True
• F. False

Answer: False

The conceptual content — mass, volume, center of mass — is identical regardless of coordinate system. Only the computational path changes. The Jacobian |J| accounts for the change in volume element, so the integral still computes the same physical quantity. Choosing spherical coordinates for a spherical region makes the calculation easier, but does not alter what is being computed.

Question 5

Explain why choosing the right coordinate system can turn a very difficult triple integral into a routine one.

Model answer: When the coordinate system matches the geometry of the region, the integration limits become constants (e.g., ρ from 0 to R for a sphere in spherical coordinates), eliminating the need to express curved boundaries as complicated functions. Additionally, the integrand often simplifies when expressed in the natural coordinates — for example, x² + y² = r² in cylindrical coordinates. The Jacobian introduces a factor (like ρ² sin φ), but this is a known function that integrates cleanly, whereas Cartesian limits for curved regions produce square-root expressions that may have no closed form.

The key principle is symmetry matching: a sphere has spherical symmetry, a cylinder has cylindrical symmetry, a box has Cartesian symmetry. When you align the coordinate axes with the natural symmetry of the region, both the limits and the integrand simplify simultaneously.