Questions: Surface Integrals and Flux of Vector Fields

Question 1

A vector field F is everywhere parallel to a flat surface S (it flows along the surface but never through it). What is the flux ∬_S F · dS?

• A. The flux equals |F| times the area of S, since F is uniformly distributed across it
• B. The flux is zero, because F has no component in the normal direction
• C. The flux is positive if F points in the direction of integration and negative otherwise
• D. The flux equals the divergence of F integrated over S

Answer: (B)

Flux measures the net flow of F *through* the surface — the component perpendicular to S. If F is parallel to S, then F · n̂ = 0 at every point: the field flows along the surface without crossing it. The dot product in the integrand extracts only the normal component, so a purely tangential field contributes exactly zero flux. Option A is the most common misconception — confusing the magnitude of the field with its ability to cross the surface.

Question 2

In the surface integral ∬_S F · dS using parametrization r(u, v), what does the cross product r_u × r_v represent?

• A. A tangent vector to the surface in the u-direction
• B. The scalar area element du dv, scaled by the parametrization
• C. A normal vector to the surface whose magnitude measures local area distortion
• D. The gradient of the scalar field associated with F

Answer: (C)

The partial derivatives r_u and r_v are tangent vectors to the surface. Their cross product r_u × r_v is perpendicular to both — pointing in the normal direction — and its magnitude |r_u × r_v| measures how much the parametrization stretches or compresses a small parameter rectangle onto the actual surface. Bundled together as dS = (r_u × r_v) du dv, it serves as a vector area element that encodes both direction and area in one quantity.

Question 3

Flipping the orientation of a surface (choosing the inward rather than outward normal) negates the entire flux integral.

• T. True
• F. False

Answer: True

Orientation is encoded in the direction of the normal vector. Swapping the normal to its opposite negates the cross product r_u × r_v, which negates every dot product F · (r_u × r_v) in the integrand. The entire integral changes sign. This is why orientation must be specified for any flux calculation — on a closed surface, the convention is the outward normal, so flux entering the surface is negative and flux leaving is positive.

Question 4

The flux integral ∬_S F · dS measures how much of the vector field F is flowing tangent to (along) the surface S.

• T. True
• F. False

Answer: False

Flux measures the component of F *perpendicular* to S — how much of the field passes through the surface. The dot product F · n̂ extracts exactly this normal component; the tangential component contributes nothing. A field flowing entirely parallel to the surface contributes zero flux. This is the essential geometric meaning: flux counts crossings, not tangential flow.

Question 5

Why does the flux integral use F · dS (the dot product with the normal vector) rather than just integrating the magnitude |F| over the surface?

Model answer: Because flux measures how much of F passes *through* the surface, not how strong F is at the surface. Only the component of F in the direction of the normal n̂ actually crosses S; the tangential component flows along the surface without contributing. The dot product F · n̂ extracts this perpendicular component. Integrating |F| would include tangential contributions and would not represent physical flux (net flow through the surface).

This distinction is critical for physical interpretation. A river flowing parallel to a mesh contributes zero net water through the mesh, even if the current is fast. Only flow in the direction the mesh is facing (the normal direction) passes through. The dot product is the mathematical operation that isolates exactly this component, which is why it appears in every flux integral.