What role does the factor ||r_u × r_v|| play in the surface integral ∬_S f dS = ∬_D f(r(u,v)) ||r_u × r_v|| du dv?
AIt gives the unit normal direction to orient the surface
BIt converts from parameter space area to actual surface area, accounting for how the parametrization stretches or compresses the surface
CIt ensures the integral converges by bounding the integrand
DIt computes the volume enclosed beneath the surface
The factor ||r_u × r_v|| is the Jacobian of the parametrization — it measures how much area in the parameter domain (du dv) corresponds to actual surface area (dS). When the surface tilts or stretches relative to parameter space (as with a hemisphere tilting toward its equator), this factor compensates. Option A describes the direction of r_u × r_v (a normal vector), not its magnitude.
Question 2 Multiple Choice
For the flat surface r(x, y) = (x, y, 0) parametrized over a region D in the xy-plane, what is ||r_x × r_y||?
A0, because the flat surface has no curvature
B1, because this parametrization maps the parameter domain to the surface without any stretching
Cx² + y², accounting for the distance from the origin
D√2, because two coordinate vectors are combined
r_x = (1, 0, 0) and r_y = (0, 1, 0), so r_x × r_y = (0, 0, 1), which has magnitude 1. A flat surface parametrized directly by its own coordinates has no stretching — the parameter area equals the surface area exactly. This makes intuitive sense: integrating f = 1 over D with dS = 1 · dx dy gives the ordinary area of D, as expected.
Question 3 True / False
The surface area element dS = ||r_u × r_v|| du dv is the 2D analogue of the arc length element ds = ||r'(t)|| dt.
TTrue
FFalse
Answer: True
The analogy is precise. For a curve, ||r'(t)|| converts from the parameter t to arc length — it is the 'speed' of the curve. For a surface, ||r_u × r_v|| converts from the 2D parameter area du dv to actual surface area dS. In both cases, the magnitude of the derivative(s) is the Jacobian that accounts for how the parametrization maps to geometry.
Question 4 True / False
The value of a surface integral ∬_S f dS depends on which parametrization of S you choose — different parametrizations give different answers.
TTrue
FFalse
Answer: False
The surface integral is a geometric quantity intrinsic to the surface S and the function f — it does not depend on the parametrization. Different parametrizations produce different integrands (different ||r_u × r_v||), but these differences exactly cancel when the domain of integration changes accordingly. The final numerical value is the same. This is analogous to how the arc length of a curve is independent of how you parametrize it.
Question 5 Short Answer
Explain why the cross product r_u × r_v appears in the surface integral formula. What geometric quantity does its magnitude represent?
Think about your answer, then reveal below.
Model answer: The partial derivatives r_u and r_v are tangent vectors to the surface in the u- and v-directions. A small rectangle [u, u+du] × [v, v+dv] in parameter space maps to a small parallelogram on the surface spanned by r_u · du and r_v · dv. The area of a parallelogram spanned by two vectors equals the magnitude of their cross product, so ||r_u × r_v|| du dv is the area of that surface patch — the infinitesimal surface area element dS.
This connects the formula back to the cross product's geometric meaning. The cross product r_u × r_v simultaneously gives the area (via its magnitude) and the normal direction (via its direction) of the surface patch. For scalar surface integrals, only the magnitude matters; for vector surface integrals (flux), the direction matters too, which is why orientation becomes significant in the next topic.