Parametric Surfaces and Surface Integrals

The line integral integrated a scalar or a field along a one-dimensional curve. The surface integral does the same over a two-dimensional surface. The step up requires a way to describe surfaces flexibly — not just graphs $z = f (x, y)$ but spheres, cylinders, and tori — and a surface area element to integrate against. Both come from parametrizing the surface with two parameters and taking the cross product of the resulting tangent vectors. Once flux, the integral of a field's normal component, is defined, Stokes' Theorem and the Divergence Theorem can be stated.

Parametric surfaces

A curve needs one parameter; a surface needs two. A parametric surface is traced by a vector function of two variables ranging over a region $D$ in the parameter plane.

A graph $z = f (x, y)$ is the special case $r (x, y) = ⟨ x, y, f (x, y)⟩$ . The sphere of radius $a$ uses the two spherical angles as parameters, $r (ϕ, θ) = ⟨ a sin ϕ cos θ, a sin ϕ sin θ, a cos ϕ ⟩$ with $0 \leq ϕ \leq π$ , $0 \leq θ \leq 2 π$ ; the grid curves are circles of latitude and meridians of longitude.

A parametric surface is meshed by two families of grid curves, one for each parameter held fixed; the tangent vectors

r_{u}

and

r_{v}

run along them.

The tangent plane and the surface area element

At a point $r (u_{0}, v_{0})$ , the two grid curves have tangent vectors given by the partial derivatives

r_{u} = \frac{\partial x}{\partial u} \overset{}{^} + \frac{\partial y}{\partial u} \overset{}{^} + \frac{\partial z}{\partial u} \hat{k}, r_{v} = \frac{\partial x}{\partial v} \overset{}{^} + \frac{\partial y}{\partial v} \overset{}{^} + \frac{\partial z}{\partial v} \hat{k} .

When $r_{u} \times r_{v} \neq = 0$ the surface is smooth there, and the two tangent vectors span the tangent plane. Their cross product $r_{u} \times r_{v}$ is normal to the surface. A small parameter rectangle of sides $Δ u$ and $Δ v$ maps to a small patch that is nearly the parallelogram spanned by $r_{u} Δ u$ and $r_{v} Δ v$ , whose area is $∣ r_{u} \times r_{v} ∣ Δ u Δ v$ .

The image of a parameter rectangle is nearly the parallelogram spanned by

r_{u} d u

and

r_{v} d v

; its area

∣ r_{u} \times r_{v} ∣ d u d v

is the surface area element.

For a graph $z = f (x, y)$ this specializes to the earlier surface-area formula: with $r (x, y) = ⟨ x, y, f ⟩$ , the cross product is $⟨ - f_{x}, - f_{y}, 1 ⟩$ and its magnitude is $f_{x}^{2} + f_{y}^{2} + 1$ .

Surface integrals of scalar functions

With the area element in hand, integrating a scalar function over a surface copies the line-integral construction.

Setting $f = 1$ recovers the surface area. With $f = ρ$ the surface density of a thin curved shell, the integral is its mass. The rule for evaluation is mechanical: substitute the parametrization into $f$ , multiply by the magnitude of the cross product, and integrate over the parameter domain — the surface analog of $\int_{C} f d s = \int f (r (t)) ∣ r^{'} (t) ∣ d t$ .

Orientation

To integrate a vector field through a surface, the surface needs a chosen side. At each point there are two unit normals, $\pm (r_{u} \times r_{v}) /∣ r_{u} \times r_{v} ∣$ ; picking one consistently orients the surface.

Most surfaces are orientable and have two sides. The Möbius band is the standard exception: sliding the normal once around returns it reversed, so no consistent choice exists, and flux through it is undefined.

An orientable surface admits a continuous field of unit normals; a closed surface is oriented outward, its normals pointing away from the enclosed solid.

Flux: the surface integral of a vector field

The physically important surface integral measures how much of a field passes through the surface. Only the component of $F$ along the normal contributes; the tangential part slides along the surface without crossing it.

In parametric form the normal and the area element combine: since $\overset{n}{^} d S = \frac{r _{u} \times r _{v}}{∣ r _{u} \times r _{v} ∣} ∣ r_{u} \times r_{v} ∣ d A = (r_{u} \times r_{v}) d A$ , the magnitude cancels and the flux is

S \iint F \cdot d S = D \iint F (r (u, v)) \cdot (r_{u} \times r_{v}) d A .

Because of the cancellation, the square root in $∣ r_{u} \times r_{v} ∣$ never appears in a flux computation. The sign depends on the orientation, which fixes whether $r_{u} \times r_{v}$ or its negative is used.

Flux counts the field's normal component crossing the surface; where

F

is perpendicular to

S

the crossing is maximal, where tangent it is zero.

The result equals the volume of the ball times $3$ ; the Divergence Theorem will explain why, since $div F = 3$ and the ball has volume $\frac{4}{3} π$ .

The three surface integrals side by side

Every surface integral here is one of these, and all three reduce to a double integral over the parameter domain $D$ .

Integral	Integrand	Reduces to	Measures
area	$1$	$D \iint ∣ r_{u} \times r_{v} ∣ d A$	area of $S$
scalar surface integral	$f$	$D \iint f (r) ∣ r_{u} \times r_{v} ∣ d A$	mass of a shell, average value
flux	$F \cdot \overset{n}{^}$	$D \iint F (r) \cdot (r_{u} \times r_{v}) d A$	flow across $S$

Flux is the surface integral that appears in Stokes' Theorem and the Divergence Theorem: the first equates a line integral around the boundary of a surface with the flux of the curl through it, the second equates the outward flux across a closed surface with the integrated divergence inside.¹

Stewart, §16.6 — Parametric Surfaces and Their Areas; §16.7 — Surface Integrals. The area element $∣ r_{u} \times r_{v} ∣ d A$ , orientation by a continuous unit normal, and the flux integral $S \iint F \cdot d S = D \iint F \cdot (r_{u} \times r_{v}) d A$ . ↩