Questions: Surface Integrals and Flux of Vector Fields

Question 1

A vector field F is tangent to every point on surface S — it lies parallel to the surface at each point. What is the flux ∬_S F · n dS?

• A. It depends on the magnitude of F
• B. It equals the surface area of S
• C. Zero, because F has no component perpendicular to S
• D. It requires knowing the parameterization to determine

Answer: (C)

Flux is F · n — the dot product projects F onto the unit normal. If F is everywhere tangent to S, it is everywhere perpendicular to n, so F · n = 0 at every point and the integral is zero. The magnitude of F is irrelevant; only the component in the normal direction contributes to flow through the surface.

Question 2

After computing r_u × r_v for a closed surface parameterization, you find the cross product points inward rather than outward. How do you correct the flux calculation?

• A. Multiply the area element |r_u × r_v| by −1
• B. Negate the entire flux integral
• C. Swap u and v in the parameterization to reverse the cross product
• D. Either negating the integral or swapping u and v achieves the correct outward flux

Answer: (D)

Both approaches work and produce the same result. Swapping u and v reverses r_u × r_v (since r_v × r_u = −(r_u × r_v)), which negates the integral. Alternatively, just multiplying the final answer by −1 is equivalent. The key point is that orientation is a choice that determines sign — if your parameterization produces the wrong normal direction, you fix it by flipping the sign of the result, not by recomputing everything.

Question 3

The flux of a vector field through a surface is generally non-negative, since it measures how much field passes through.

• T. True
• F. False

Answer: False

Flux is a signed quantity. The sign depends on the orientation: if F has a component opposing the chosen normal direction n, the dot product F · n is negative at those points, and the integral can be negative. Negative flux simply means net flow in the direction opposite to n. For example, outward flux through a closed surface can be negative if the field converges inward more than it diverges outward.

Question 4

For an incompressible fluid with velocity field F satisfying ∇·F = 0, the net flux of F through any closed surface is zero.

• T. True
• F. False

Answer: True

By the Divergence Theorem, the net outward flux through a closed surface S equals ∭_V ∇·F dV over the enclosed volume. If ∇·F = 0 everywhere inside, this triple integral is zero. Physically, an incompressible fluid has no sources or sinks — exactly as much fluid flows into any closed region as flows out of it, so the net flux is zero.

Question 5

Why does the orientation of a surface (the choice of normal direction) matter when computing flux, and what determines which direction the normal points?

Model answer: Orientation determines the sign of the flux. Flux measures flow through a surface in a specific direction — positive when F has a component in the direction of n, negative when it opposes n. The normal direction is set by the parameterization: the cross product r_u × r_v points in one of two possible normal directions, and swapping the order u ↔ v reverses it. For closed surfaces, the convention is the outward normal (pointing away from the enclosed region). After computing r_u × r_v, you verify it points the intended way — if not, negate the integral. Getting orientation wrong doesn't change the magnitude; it flips the sign.

This is not merely a formality. In the Divergence Theorem, orientation must be consistent (outward for the closed surface) for the theorem to hold. In Stokes' Theorem, the boundary orientation and surface normal are linked by the right-hand rule. Every flux computation requires specifying orientation explicitly, because without it the integral is ambiguous in sign.