A thin metal shell has density (mass per unit area) given by ρ(x, y, z). Which integral computes the total mass of the shell?
A∭_V ρ dV, integrating density over the volume enclosed by the shell
B∬_S ρ dS, integrating density over the surface using the area element dS
C∬_D ρ(r(u,v)) du dv, integrating over the parameter domain without correction
D∮_C ρ ds, integrating density along a boundary curve of the surface
A thin shell has no volume — its mass is distributed over a 2D surface. The correct integral is ∬_S ρ dS, where dS = |r_u × r_v| du dv is the area element that accounts for the curvature of the surface. Option C is the most tempting error: integrating ρ over the flat parameter domain without the |r_u × r_v| factor ignores the area magnification, giving an incorrect answer whenever the surface is curved or the parametrization is non-unit-speed. This is the surface analog of computing mass of a rod as ∫_a^b ρ(x) dx.
Question 2 Multiple Choice
What role does the factor |r_u × r_v| play in the surface integral formula ∬_S f dS = ∬_D f(r(u,v)) |r_u × r_v| du dv?
AIt normalizes the integrand so the result is independent of units
BIt converts flat parameter-domain area du dv into actual curved surface area, acting as an area magnification factor
CIt gives the direction of the surface normal, which determines the sign of the integral
DIt cancels out when f = 1, leaving just the area of the parameter domain
|r_u × r_v| is the area magnification: it measures how much a small rectangle du dv in the parameter domain gets stretched into a surface patch on the actual curved surface. This factor is what makes dS a genuine geometric quantity — independent of parametrization — rather than just du dv. When f = 1, the formula reduces to ∬_D |r_u × r_v| du dv, which is the surface area formula (not the area of the parameter domain, which would be wrong unless the surface happens to be flat and unit-speed parametrized). Note that |r_u × r_v| gives magnitude only; the next topic introduces the signed vector element r_u × r_v for flux integrals.
Question 3 True / False
Two different valid parametrizations of the same surface will give the same value for ∬_S f dS.
TTrue
FFalse
Answer: True
dS is a geometric quantity attached to the surface — it measures actual area, not parameter-domain area. Because |r_u × r_v| automatically adjusts for how the parametrization stretches the surface, any valid parametrization produces the same value for the integral. This is the surface analog of parametrization-independence for line integrals: ∫_C f ds gives the same result regardless of how you parametrize the curve, because the speed |r'(t)| compensates exactly.
Question 4 True / False
The surface integral ∬_S f dS depends on which parametrization r(u, v) you choose — different parametrizations may give different numerical results.
TTrue
FFalse
Answer: False
This is the most common conceptual error. The factor |r_u × r_v| is precisely what makes the integral parametrization-independent: it converts the flat (u,v) area into actual surface area, compensating for any stretching or compression introduced by the particular parametrization chosen. The integral ∬_S f dS is a geometric quantity — it depends on the function f and the surface S, not on the coordinate system used to describe S.
Question 5 Short Answer
Explain what the factor |r_u × r_v| represents geometrically and why it must appear in the surface integral formula.
Think about your answer, then reveal below.
Model answer: |r_u × r_v| is the area magnification factor: its magnitude equals the area of the parallelogram spanned by the tangent vectors r_u and r_v, which represents how much a small unit patch in the parameter domain gets stretched into actual surface area. It must appear because different regions of a curved surface may be compressed or expanded relative to the flat parameter domain, and the integral must account for the true area of each surface patch — not the area of the corresponding patch in (u,v) space.
Without |r_u × r_v|, integrating over the parameter domain would give the correct answer only if the parametrization happened to be isometric (area-preserving), which is rare. The cross product r_u × r_v has direction (the surface normal) and magnitude (area of the tangent parallelogram), making it the natural object to represent an infinitesimal oriented surface patch. For scalar integrals we only need the magnitude; for vector (flux) integrals we also use the direction.