Surface integrals — scalar and vector (flux): Question bank

The core objects you'll be tested on:

The stretch factor $∥ r_{u} \times r_{v} ∥$ — how much area a tiny parameter square gains when mapped onto the curved surface (built from the Cross product).
The scalar element $d S = ∥ r_{u} \times r_{v} ∥ d u d v$ (always positive).
The vector element $d S = (r_{u} \times r_{v}) d u d v$ (carries a sign / direction).

True or false — justify

Cover the answers. Each reveal is a full reason.

Flux

\iint_{S} F \cdot d S

is always

\geq 0

when

F

has positive components.

False. Flux measures flow along the chosen normal; if the field points opposite the normal on part of the surface, that part contributes negatively, so the total can be any sign.

A scalar surface integral

\iint_{S} f d S

with

f \geq 0

is always

\geq 0

True. Here

d S = ∥ r_{u} \times r_{v} ∥ d u d v

is a magnitude (always

\geq 0

), and there is no normal direction to carry a sign, so a nonnegative integrand gives a nonnegative result.

Reversing the parametrization (swapping

u

and

v

) changes the value of a scalar surface integral.

False. Swapping

u, v

replaces

r_{u} \times r_{v}

r_{v} \times r_{u} = - (r_{u} \times r_{v})

, but

d S

uses the magnitude, which is unchanged; only flux (which keeps the sign) flips.

The stretch factor

∥ r_{u} \times r_{v} ∥

is the 3D analogue of the Jacobian used in Double integrals.

True. Both convert a tiny parameter-region area into the actual area it covers after the map; the Jacobian does it inside a plane,

∥ r_{u} \times r_{v} ∥

does it for a plane mapped into 3D space.

For a flux integral you must always compute the square root

∥ r_{u} \times r_{v} ∥

False. In

\iint (F \cdot \hat{n}) d S

the

∥ \dots ∥

\hat{n}

cancels the

∥ \dots ∥

d S

, leaving

F \cdot (r_{u} \times r_{v}) d u d v

— no root needed.

Every smooth closed surface is orientable.

True. A smooth closed surface (like a sphere) has a consistent inside/outside, so you can pick "outward" everywhere; non-orientable surfaces like the Möbius strip are the exceptions and are not closed.

If two different parametrizations give the same oriented surface, they give the same flux.

True. Flux depends only on the surface, the field, and the chosen normal direction — not on how you sweep the parameters — as long as the orientation (normal direction) agrees.

The vector

r_{u} \times r_{v}

is always a unit vector.

False. Its length is the stretch factor

∥ r_{u} \times r_{v} ∥

, which is generally not

1

; you must divide by that length to get the unit normal

\hat{n}

For the graph

z = g (x, y)

, the element

r_{x} \times r_{y} = (- g_{x}, - g_{y}, 1)

always points upward.

True. Its

z

-component is

+ 1 > 0

regardless of

g

, so this specific parametrization always yields the upward-pointing normal.

Spot the error

Each line states a flawed move; the reveal names the mistake and fixes it.

"Since

F = (x, y, z)

and on the sphere

r

is the outward normal, I'll use

F \cdot r

for flux."

The vector area element is

r_{ϕ} \times r_{θ}

, not

r

. It only happens to be parallel to

r

on the sphere; the correct object is always the cross product of the tangent vectors.

\iint_{S} f d S = \iint_{D} f (x, y, z) d u d v

Two errors: the

∥ r_{u} \times r_{v} ∥

stretch factor is missing, and

f

must be rewritten as

f (r (u, v))

— everything must become a function of

u, v

before integrating.

"To find the mass of a curved sheet with density

f

, integrate

f

over the flat shadow region

D

You forgot the stretch: slanted patches cover more real area than their shadow, so mass

= \iint_{D} f ∥ r_{u} \times r_{v} ∥ d u d v

, not

\iint_{D} f d u d v

"The flux came out negative, so I made an arithmetic mistake."

Not necessarily — a negative flux just means net flow is opposite your chosen normal. It's only wrong if it contradicts the orientation the problem asked for.

"For flux through

z = g (x, y)

I'll use

\iint_{D} (P + Q + R) d x d y

You dropped the slope weighting. Dotting

F = (P, Q, R)

with

(- g_{x}, - g_{y}, 1)

gives

\iint_{D} (- P g_{x} - Q g_{y} + R) d x d y

"I checked with the Divergence Theorem but the surface is a hemisphere with an open bottom, so flux

= ∭ \nabla \cdot F d V

The Divergence Theorem needs a closed surface. You must add the flux through the flat disk that caps the hemisphere before equating to the volume integral.

"I parametrized the surface, but plugged the original

x, y, z

into

F

and integrated over

D

F

must be evaluated at

r (u, v)

, i.e.

F (x (u, v), y (u, v), z (u, v))

, so the integrand is a genuine function of

u, v

over

D

Why questions

Why does the area of the mapped parallelogram equal

∥ r_{u} \times r_{v} ∥

rather than

∥ r_{u} ∥ \cdot ∥ r_{v} ∥

Because the tangent vectors need not be perpendicular;

∥ r_{u} \times r_{v} ∥ = ∥ r_{u} ∥∥ r_{v} ∥ sin θ

accounts for the angle between them (base times height), while a plain product would overcount.

Why does only the component of

F

along

\hat{n}

contribute to flux?

Flow parallel to the surface slides across it without passing through; only the perpendicular part actually crosses, and

F \cdot \hat{n}

extracts exactly that perpendicular component.

Why is flux often easier to compute than a scalar surface integral over the same surface?

The scalar integral keeps the awkward square root

∥ r_{u} \times r_{v} ∥

, whereas in flux that norm cancels, leaving a plain polynomial-like dot product

F \cdot (r_{u} \times r_{v})

Why do we use the Cross product here instead of, say, the dot product to build

d S

We need the area spanned by two vectors and a direction perpendicular to both; the cross product delivers both (its magnitude is the area, its direction is the normal), which the dot product cannot.

Why does choosing

\hat{n}

versus

- \hat{n}

matter for flux but not for a scalar integral?

Flux is a signed directional quantity ("through which way?"), so the normal's direction sets the sign; a scalar integral only sums positive area contributions and has no direction to reverse.

Why does the graph case give

∥ r_{x} \times r_{y} ∥ = 1 + g_{x}^{2} + g_{y}^{2}

, and what does the extra

\geq 1

mean geometrically?

The

1

is the flat shadow contribution and

g_{x}^{2} + g_{y}^{2}

adds the tilt; being

\geq 1

means a sloped patch is always larger than its horizontal shadow, never smaller.

Why can a Möbius strip have no consistent orientation?

Sliding the normal once around the strip returns it pointing the opposite way, so no continuous global choice of

\hat{n}

exists — the surface has only "one side."

Edge cases

F

is everywhere tangent to the surface, what is the flux?

Exactly zero, because

F \cdot \hat{n} = 0

at every point — nothing crosses through, everything slides along.

What does

\iint_{S} f d S

give when

f \equiv 1

The surface area itself, since you are just summing the area of every patch with no weighting.

At the equator of a sphere parametrized by

(ϕ, θ)

ϕ = π /2

; what special role does

sin ϕ

play at the poles

ϕ = 0, π

The spherical area factor

sin ϕ

vanishes at the poles, correctly reflecting that the parameter grid squares collapse to zero area there (a coordinate degeneracy, not a hole).

If a surface is only piecewise smooth (like a cube with edges), can you still integrate over it?

Yes — split it into smooth faces, integrate each separately, and add; the edges are measure-zero and contribute nothing, but you must keep each face's orientation consistent (all outward) for flux.

What happens to flux if the surface encloses no net source and

\nabla \cdot F = 0

inside a closed region?

By the Divergence Theorem the total outward flux is zero — as much fluid enters as leaves, so the field is "incompressible" across that closed surface.

∥ r_{u} \times r_{v} ∥

ever zero, and what does that signal?

Yes, wherever

r_{u}

and

r_{v}

are parallel (or one is zero); the parametrization degenerates there (like the poles above), producing no area — a defect of the chart, not the surface.

How does a surface integral relate to a line integral via Stokes' Theorem?

Stokes' Theorem equates the flux of

\nabla \times F

through a surface to the line integral of

F

around its boundary curve, linking the 2D surface flux to a 1D loop.

Recall One-line summaries to carry away

Scalar $d S$ : magnitude, always positive, keeps the root.
Vector $d S$ : signed, direction $= \hat{n}$ , root cancels.
Reverse orientation → flux flips sign, scalar unchanged.
Tangent field → zero flux; closed surface with $\nabla \cdot F = 0$ → zero net flux.