Questions: Applications of Double Integrals: Area, Volume, and Mass

Question 1

A thin plate (lamina) occupies the region R with area 6 m² and has uniform density ρ = 4 kg/m². What is its mass, and which double integral computes it?

• A. Mass = 1.5 kg, computed by ∬_R (1/ρ) dA
• B. Mass = 10 kg, computed by ∬_R (ρ + 1) dA
• C. Mass = 24 kg, computed by ∬_R ρ dA = ρ · Area(R)
• D. Mass = 6 kg, computed by ∬_R 1 dA regardless of density

Answer: (C)

Mass equals ∬_R ρ(x,y) dA. When density is uniform (constant ρ), the integral factors: ρ∬_R 1 dA = ρ · Area(R) = 4 · 6 = 24 kg. Each infinitesimal area element dA has mass ρ · dA, and summing over the plate gives total mass. Option D correctly notes that ∬_R 1 dA = Area(R) = 6, but ignores the density factor. The formula ∬_R ρ dA reduces to ρ · Area only for uniform density — for variable density you must integrate ρ(x,y) directly.

Question 2

You want the volume under the paraboloid z = x² + y² above the unit disk R: x² + y² ≤ 1. Which setup is correct?

• A. ∬_R 1 dA — the unit disk has area π, giving volume π
• B. ∬_R (x² + y²) dA — integrate the surface height over the region
• C. ∫₀¹ ∫₀¹ (x² + y²) dy dx — integrate over the unit square
• D. ∫₀¹ (x² + y²)² dx — square the integrand to account for the 2D region

Answer: (B)

Volume under z = f(x,y) above region R is ∬_R f(x,y) dA, where each infinitesimal area element dA supports a column of height f(x,y). Here f = x² + y², so the setup is ∬_R (x² + y²) dA over the unit disk. Option A gives the area of R (= π), not the volume. Option C integrates over the wrong region (a square, not a disk). This integral is most naturally evaluated in polar coordinates: ∫₀²π ∫₀¹ r² · r dr dθ.

Question 3

The integral ∬_R 1 dA generally equals 1, regardless of the shape or size of the region R.

• T. True
• F. False

Answer: False

∬_R 1 dA gives the area of R. When f(x,y) = 1, each infinitesimal area element dA contributes exactly dA to the integral — summing these up gives the total area of the region. A unit square gives 1, a circle of radius 2 gives 4π, and so on. The result is 1 only if R happens to have area 1.

Question 4

For a non-uniform lamina with density ρ(x,y), the formula ∬_R ρ(x,y) dA reduces to ρ · Area(R) only when the density is constant.

• T. True
• F. False

Answer: True

When ρ is constant, it factors out of the integral: ∬_R ρ dA = ρ ∬_R 1 dA = ρ · Area(R). When ρ varies with position, different patches of the lamina contribute different amounts of mass per unit area, and you must integrate ρ(x,y) weighted by dA across the whole region. The constant-density formula is a special case, not the general rule.

Question 5

Explain why the formulas for volume under a surface and mass of a lamina have identical mathematical structure, even though they describe physically different quantities.

Model answer: Both are double integrals of the form ∬_R f(x,y) dA. In both cases, the integrand assigns a 'weight' to each infinitesimal area element dA: for volume, f(x,y) is the height of the surface above that patch, so f·dA is the volume of a thin column; for mass, ρ(x,y) is the mass per unit area at that patch, so ρ·dA is the mass of a tiny piece of the lamina. In both cases the double integral sums infinitely many infinitesimal contributions over the region. The physics differs (geometry vs. mass), but the mathematical operation — integrating a function over a 2D region — is identical.

This structural unity is the payoff of the integral interpretation: once you understand that ∬ f dA sums f-weighted area contributions, you can apply it to any quantity expressible as a density — mass density, probability density, charge density, and more. The application changes; the setup procedure does not.