Parametric surfaces — tangent planes, surface area: Question bank

Before we start, the words and symbols we keep reusing, in plain language:

Definition Quick vocabulary anchor

Parametric map $r (u, v) = ⟨ x (u, v), y (u, v), z (u, v)⟩$ : a rule that takes two numbers $(u, v)$ and outputs a point in 3D space. As $(u, v)$ range over their allowed values, the output point sweeps out the whole surface — think of $r$ as painting the rubber sheet using $(u, v)$ as its two colours.
Parameter region $D$ : the flat, 2D region of allowed $(u, v)$ values (a patch of the flat $uv$ -plane). It's the "flat blanket" before it's pressed onto the surface; every point of the surface comes from exactly one point of $D$ .
Tangent vector $r_{u}$ : the "velocity arrow" you get by walking along the surface while only the parameter $u$ changes. It lies flat in the surface, like an arrow drawn on the rubber sheet.
Normal vector $n = r_{u} \times r_{v}$ : an arrow sticking straight out of the surface, perpendicular to every arrow drawn on it.
Cross product $a \times b$ : a machine that eats two arrows and spits out a third arrow, perpendicular to both, whose length equals the area of the parallelogram those two arrows fence off. See Cross Product.
Dot product $a \cdot b = ∣ a ∣∣ b ∣ cos θ$ : a machine that eats two arrows and outputs a single number measuring how much they point the same way (their alignment). It is zero exactly when the arrows are perpendicular.

True or false — justify

The map

r (u, v)

only tells you where surface points are, so the surface's shape is all that matters.

False — the same set of points can be painted by different

r

's, and

r_{u}, r_{v}

depend on the painting, but the surface area

\iint_{D} ∣ r_{u} \times r_{v} ∣ d A

comes out the same because it measures the true geometric spreading.

r_{u} \times r_{v} = 0

at a point, the surface is flat there.

False — it means the parametrization is degenerate (the two grid directions collapsed or lined up), so the tangent plane isn't defined there. A perfectly smooth sphere has this happen at its poles purely because of the coordinates, not the geometry.

∣ r_{u} \times r_{v} ∣ = ∣ r_{u} ∣ ∣ r_{v} ∣

always.

False — that only holds when the two tangent arrows are perpendicular. In general

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

, which is smaller whenever the grid lines meet at an oblique angle.

Swapping the order to

r_{v} \times r_{u}

changes the tangent plane you compute.

False — it flips the normal to point the other way, but the plane through

P_{0}

perpendicular to that line is identical. Sign only matters for orientation/flux (see Surface Integrals and Flux).

The tangent plane's normal

r_{u} \times r_{v}

and the surface-area factor

∣ r_{u} \times r_{v} ∣

are unrelated quantities.

False — they are the same cross product; the normal is the vector, the area factor is its length. That's the entire "one cross does both" idea.

For a graph

z = f (x, y)

, the surface area equals the area of its shadow

D

in the

x y

-plane.

False — the shadow is

\iint_{D} d A

, but the tilted surface is longer, giving

\iint_{D} f_{x}^{2} + f_{y}^{2} + 1 d A \geq area (D)

, with equality only when the surface is horizontal (

f_{x} = f_{y} = 0

If a surface is smooth everywhere, its parametrization is smooth everywhere.

False — smoothness of the shape is geometric, but

r_{u} \times r_{v}

can still vanish at "coordinate singularities" (like the sphere's poles) where the map degenerates even though the surface is locally as smooth as anywhere else.

The area factor

∣ r_{u} \times r_{v} ∣

is a constant you can pull out of the integral.

False in general — it usually depends on

(u, v)

(e.g.

R^{2} sin ϕ

for the sphere). It only pulls out when it happens to be constant, as for a flat plane like

z = 2 x + 2 y

Spot the error

"Area

= \iint_{D} r_{u} \cdot r_{v} d A

The dot product measures alignment (

cos θ

), and it's zero exactly when the edges are perpendicular — precisely the case of maximum area. Area needs

∣ r_{u} \times r_{v} ∣

, whose factor

sin θ

peaks at

9 0^{\circ}

"Area

= \iint_{D} d A

over the parameter region, because

d A

already is area."

That's area in the flat

(u, v)

plane

D

, not on the curved surface. The map stretches each tiny square, so you must weight

d A

by the local stretch

∣ r_{u} \times r_{v} ∣

(the Jacobian-type factor, cf. Change of Variables and the Jacobian).

"The normal to

z = f (x, y)

⟨ f_{x}, f_{y}, 1 ⟩

The correct normal from

r_{x} \times r_{y}

⟨ - f_{x}, - f_{y}, 1 ⟩

— the horizontal components carry a minus sign because the surface rises against the direction of steepest increase. (For the plane equation either sign works, but the vector is

⟨ - f_{x}, - f_{y}, 1 ⟩

"For the sphere,

∣ r_{ϕ} \times r_{θ} ∣ = R^{2}

, so

A = R^{2} \cdot (π) (2 π) = 2 π^{2} R^{2}

The factor is

R^{2} sin ϕ

, not

R^{2}

— it vanishes at the poles and peaks at the equator. Integrating

R^{2} sin ϕ

gives the correct

4 π R^{2}

r_{u}

is perpendicular to the surface because it's a derivative."

The opposite —

r_{u}

is tangent: it's the velocity of a curve that lives on the surface (freeze

v

, move

u

), so it lies flat in the surface. See Tangent Lines and Velocity Vectors.

"To find the tangent plane I only need the point

P_{0}

A plane needs a point and a direction to face — the normal

n = r_{u} \times r_{v}

. Without

n

infinitely many planes pass through

P_{0}

"A parametric surface needs one parameter, like a parametric curve."

A surface is intrinsically two-dimensional — you can wander in two independent directions — so it needs two parameters

(u, v)

. One parameter only traces a curve.

Why questions

Why a cross product and not a dot product for the area factor?

Area measures how much two edge-arrows spread apart, which is

∣ r_{u} ∣∣ r_{v} ∣ sin θ

— exactly the cross-product magnitude. The dot product measures how much they align (

cos θ

), the wrong quantity.

Why does moving

Δ u

in the parameter shift the surface point by

\approx r_{u} Δ u

This is the linear (tangent-line) approximation: near a point the curve

u \mapsto r (u, v_{0})

moves at velocity

r_{u}

, so a small step

Δ u

produces displacement

\approx r_{u} Δ u

, just like distance

\approx

speed

\times

time.

Why does the little curved patch behave like a flat parallelogram?

Over an infinitesimally small

Δ u \times Δ v

the curvature is negligible and the two edges are straight arrows

r_{u} Δ u

r_{v} Δ v

. Two arrows fence off a parallelogram, whose area is a cross-product magnitude — this is why

∣ r_{u} \times r_{v} ∣

appears.

Why do we take the magnitude of the cross product for area but keep the vector for the tangent plane?

Area is a size (a non-negative number), so we need only the length; the plane needs a direction to be perpendicular to, so we keep the full vector. Same cross product, two different pieces of information used.

Why must

r_{u} \times r_{v} \neq = 0

for a "smooth" (regular) point?

If it's zero, the two tangent arrows are parallel (or one vanished), so they don't span a genuine 2D plane — there's no well-defined tangent plane and no reliable area factor there.

Why is the surface-area factor the same idea as a Jacobian in a change of variables?

Both answer "by how much does this map stretch tiny areas?" The Jacobian

∣ det J ∣

does it for maps between flat regions;

∣ r_{u} \times r_{v} ∣

does it for a map from a flat parameter region

D

onto a curved surface. See Change of Variables and the Jacobian.

Edge cases

What happens to

∣ r_{ϕ} \times r_{θ} ∣ = R^{2} sin ϕ

at the sphere's poles

ϕ = 0, π

It becomes

0

because

sin (0) = sin (π) = 0

— the

θ

-circle shrinks to a point there, so those meridians collapse. It's a coordinate singularity, not a hole in the sphere, and it contributes zero to the integral harmlessly.

If the two grid curves meet at a

0^{\circ}

angle (tangent arrows parallel), what is the patch area?

Zero, since

sin 0 = 0

: parallel arrows fence off a degenerate parallelogram with no area. The parametrization has failed to spread out in two independent directions there.

For a perfectly flat surface like

z = 2 x + 2 y

, why is the area factor constant?

f_{x} = f_{y} = 2

are constants, so

f_{x}^{2} + f_{y}^{2} + 1 = 9 = 3

everywhere; the tilt never changes. A constant factor just multiplies the shadow's area, giving

3 \cdot area (D)

What is the area factor when the surface is the flat

x y

-plane,

z = 0

f_{x} = f_{y} = 0

gives

0 + 0 + 1 = 1

, so the surface area equals the shadow area exactly — no stretching, which is the sanity check the formula must pass.

If you reverse the parameter direction (

u \to - u

), does the computed area change?

No —

r_{u}

flips sign, but

∣ r_{u} \times r_{v} ∣

uses magnitude, so the area is unchanged. Only orientation-sensitive quantities like flux would feel the flip.

What does a single tangent vector alone tell you about the surface direction — is it enough to build the plane?

No — one arrow gives only one direction in the surface. You need two non-parallel tangent arrows to span the full 2D tangent plane and to define the normal via their cross product.

Recall One-line summary of the whole trap set

Tangent arrows lie in the surface; their cross product is the normal (vector) and the area stretch (its length); use magnitude for area, keep the vector for planes, and watch for coordinate singularities where the cross product harmlessly vanishes.