---
title: Parametric Surfaces and Surface Integrals
module: Multiple Integrals and Vector Calculus
moduleNumber: 12
lessonNumber: 5
order: 1205
summary: >
  A parametric surface is the image of a two-variable vector function; its area
  element is the magnitude of the cross product of the two tangent vectors. The
  surface integral of a scalar function sums it over that area, and the flux
  integral of a vector field sums the field's normal component, measuring flow
  through the surface. Orientation by a choice of unit normal makes flux
  well-defined, the integral Stokes' and the Divergence Theorem operate on.
topics: [Multiple Integrals and Vector Calculus]
draft: false
sources:
  - book: Stewart
    ref: "Ch. 16 — Vector Calculus; §16.6 Parametric Surfaces and Their Areas"
  - book: Stewart
    ref: "§16.7 Surface Integrals"
---

The [line integral](/calculus/multiple-integrals-and-vector-calculus/vector-fields-and-line-integrals)
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.

> **Definition (Parametric surface).** A **parametric surface** is the image of a
> vector function
> $$
> \vec{r}(u,v) = x(u,v)\,\hat\imath + y(u,v)\,\hat\jmath + z(u,v)\,\hat{k}, \qquad (u,v) \in D .
> $$
> Holding $v$ fixed and varying $u$ traces a **grid curve** on the surface; holding
> $u$ fixed and varying $v$ traces another. The two families of grid curves mesh
> the surface.

A graph $z = f(x,y)$ is the special case $\vec{r}(x,y) = \langle x, y, f(x,y)\rangle$.
The sphere of radius $a$ uses the two spherical angles as parameters,
$\vec{r}(\phi,\theta) = \langle a\sin\phi\cos\theta,\ a\sin\phi\sin\theta,\ a\cos\phi\rangle$
with $0 \le \phi \le \pi$, $0 \le \theta \le 2\pi$; the grid curves are circles of
latitude and meridians of longitude.

$$
% caption: A parametric surface is meshed by two families of grid curves, one for
% each parameter held fixed; the tangent vectors $\mathbf{r}_u$ and
% $\mathbf{r}_v$ run along them.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% a curved patch drawn with two families of grid curves
\foreach \v in {0,1,2,3} {
  \draw[black] (0.0+0.25*\v, 0.6+0.55*\v) .. controls (1.8+0.2*\v,1.6+0.4*\v) and (3.4+0.15*\v,1.1+0.45*\v) .. (5.0+0.1*\v,1.7+0.4*\v);
}
\foreach \u in {0,1,2,3,4} {
  \draw[black] (0.0+1.25*\u*0.8, 0.6+0.14*\u) .. controls ({0.6+1.0*\u*0.8},{1.4+0.1*\u}) .. ({0.75+0.8*\u*0.8},{2.25+0.05*\u});
}
% tangent vectors at a point
\coordinate (P) at (2.5,1.75);
\fill[acc] (P) circle (2pt);
\draw[acc, very thick, ->] (P) -- ++(1.3,0.25);
\node[acc, font=\small, anchor=west] at (3.8,2.0) {$\mathbf{r}_u$};
\draw[acc, very thick, ->] (P) -- ++(0.25,1.1);
\node[acc, font=\small, anchor=south] at (2.75,2.85) {$\mathbf{r}_v$};
\node[font=\small, anchor=north east] at (5.1,1.7) {surface $\mathbf{r}$};
\end{tikzpicture}
$$

## The tangent plane and the surface area element

At a point $\vec{r}(u_0, v_0)$, the two grid curves have tangent vectors given
by the partial derivatives

$$
\vec{r}_u = \frac{\partial x}{\partial u}\hat\imath + \frac{\partial y}{\partial u}\hat\jmath + \frac{\partial z}{\partial u}\hat{k}, \qquad \vec{r}_v = \frac{\partial x}{\partial v}\hat\imath + \frac{\partial y}{\partial v}\hat\jmath + \frac{\partial z}{\partial v}\hat{k} .
$$

When $\vec{r}_u \times \vec{r}_v \neq \vec{0}$ the surface is **smooth**
there, and the two tangent vectors span the tangent plane. Their
[cross product](/calculus/vectors-and-space-curves/the-cross-product-lines-and-planes)
$\vec{r}_u \times \vec{r}_v$ is normal to the surface. A small parameter
rectangle of sides $\Delta u$ and $\Delta v$ maps to a small patch that is nearly
the parallelogram spanned by $\vec{r}_u\,\Delta u$ and $\vec{r}_v\,\Delta v$,
whose area is $|\vec{r}_u \times \vec{r}_v|\,\Delta u\,\Delta v$.

$$
% caption: The image of a parameter rectangle is nearly the parallelogram spanned
% by $\mathbf{r}_u\,du$ and $\mathbf{r}_v\,dv$; its area
% $|\mathbf{r}_u\times\mathbf{r}_v|\,du\,dv$ is the surface area element.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% parameter rectangle (left)
\draw[->, black] (-0.3,0) -- (2.4,0) node[right, font=\small] {$u$};
\draw[->, black] (0,-0.3) -- (0,2.3) node[above, font=\small] {$v$};
\fill[black!8] (0.7,0.6) rectangle (1.5,1.4);
\draw[thick] (0.7,0.6) rectangle (1.5,1.4);
\node[font=\small] at (1.1,1.0) {$du\,dv$};
\node[font=\small, anchor=north] at (1.1,-0.35) {parameter domain};
% arrow
\draw[->, very thick] (2.8,1.1) -- (4.1,1.1);
\node[font=\small, anchor=south] at (3.45,1.15) {$\mathbf{r}$};
% surface parallelogram (right)
\begin{scope}[xshift=4.7cm, yshift=0.2cm]
  \fill[acc!12] (0.5,0.5) -- (2.0,0.9) -- (2.4,1.9) -- (0.9,1.5) -- cycle;
  \draw[acc, thick] (0.5,0.5) -- (2.0,0.9) -- (2.4,1.9) -- (0.9,1.5) -- cycle;
  \draw[acc, very thick, ->] (0.5,0.5) -- (2.0,0.9);
  \node[acc, font=\small, anchor=north] at (1.4,0.6) {$\mathbf{r}_u\,du$};
  \draw[acc, very thick, ->] (0.5,0.5) -- (0.9,1.5);
  \node[acc, font=\small, anchor=east] at (0.55,1.1) {$\mathbf{r}_v\,dv$};
  \node[font=\small, anchor=north] at (1.4,-0.2) {surface patch $dS$};
\end{scope}
\end{tikzpicture}
$$

> **Definition (Surface area).** For a smooth parametric surface $S$ covered once
> by $\vec{r}(u,v)$ over $D$, the **surface area element** is
> $\d S = |\vec{r}_u \times \vec{r}_v|\,\d A$, and the area is
> $$
> A(S) = \iint\limits_D |\vec{r}_u \times \vec{r}_v|\,\d A .
> $$

For a graph $z = f(x,y)$ this specializes to the earlier
[surface-area formula](/calculus/multiple-integrals-and-vector-calculus/triple-integrals-and-coordinate-systems):
with $\vec{r}(x,y) = \langle x, y, f\rangle$, the cross product is
$\langle -f_x, -f_y, 1\rangle$ and its magnitude is $\sqrt{f_x^2 + f_y^2 + 1}$.

> **Worked example.** The surface area of a sphere of radius $a$. With
> $\vec{r}(\phi,\theta)$ as above, a computation gives
> $|\vec{r}_\phi \times \vec{r}_\theta| = a^2\sin\phi$, so
>
> $$
> A = \int_0^{2\pi}\int_0^\pi a^2\sin\phi\,\d\phi\,\d\theta = a^2 (2\pi)\big[-\cos\phi\big]_0^\pi = a^2(2\pi)(2) = 4\pi a^2 .
> $$

## Surface integrals of scalar functions

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

> **Definition (Surface integral of a scalar function).** For $f$ continuous on a
> smooth surface $S$ parametrized by $\vec{r}(u,v)$ over $D$,
> $$
> \iint\limits_S f(x,y,z)\,\d S = \iint\limits_D f(\vec{r}(u,v))\,|\vec{r}_u \times \vec{r}_v|\,\d A .
> $$

Setting $f = 1$ recovers the surface area. With $f = \rho$ 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
$\int_C f\,\d s = \int f(\vec{r}(t))|\vec{r}\,'(t)|\,\d t$.

> **Worked example.** Evaluate $\iint\limits_S z\,\d S$ where $S$ is the part of the plane
> $z = 1 + 2x + 3y$ above the rectangle $[0,1]\times[0,2]$. As a graph,
> $|\vec{r}_x \times \vec{r}_y| = \sqrt{4 + 9 + 1} = \sqrt{14}$, so
>
> $$
> \iint\limits_S z\,\d S = \int_0^1 \int_0^2 (1 + 2x + 3y)\sqrt{14}\,\d y\,\d x = \sqrt{14}\int_0^1 \big[y + 2xy + \tfrac{3y^2}{2}\big]_0^2 \d x .
> $$
>
> The inner integral is $2 + 4x + 6 = 8 + 4x$, so the value is
> $\sqrt{14}\int_0^1 (8 + 4x)\,\d x = \sqrt{14}\,(8 + 2) = 10\sqrt{14}$.

## Orientation

To integrate a vector field through a surface, the surface needs a chosen side.
At each point there are two unit normals, $\pm(\vec{r}_u\times\vec{r}_v)/|\vec{r}_u\times\vec{r}_v|$;
picking one consistently orients the surface.

> **Definition (Oriented surface).** A surface is **orientable** if a continuous
> choice of unit normal $\hat{n}$ can be made over the whole surface. Such a
> choice is an **orientation**. A closed surface is conventionally oriented by the
> **outward** normal, pointing away from the enclosed solid.

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.

$$
% caption: An orientable surface admits a continuous field of unit normals; a
% closed surface is oriented outward, its normals pointing away from the enclosed
% solid.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% a curved open surface
\draw[thick] (0.2,0.6) .. controls (1.6,2.0) and (3.4,-0.2) .. (4.8,1.2);
\draw[thick] (0.2,1.6) .. controls (1.6,3.0) and (3.4,0.8) .. (4.8,2.2);
\draw[thick] (0.2,0.6) -- (0.2,1.6);
\draw[thick] (4.8,1.2) -- (4.8,2.2);
% consistent normals
\foreach \x/\y/\dx/\dy in {1.1/1.7/-0.2/0.7, 2.5/1.15/0.05/0.75, 3.9/1.55/0.25/0.7} {
  \draw[acc, thick, ->] (\x,\y) -- ({\x+\dx},{\y+\dy});
}
\node[acc, font=\small, anchor=south] at (2.55,1.95) {$\mathbf{n}$};
\node[font=\small, anchor=north] at (2.5,0.3) {consistent unit normals};
\end{tikzpicture}
$$

## 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 $\vec{F}$ along the normal
contributes; the tangential part slides along the surface without crossing it.

> **Definition (Flux).** The **flux** of a vector field $\vec{F}$ across an
> oriented surface $S$ with unit normal $\hat{n}$ is
> $$
> \iint\limits_S \vec{F}\cdot \d\vec{S} = \iint\limits_S \vec{F}\cdot\hat{n}\,\d S .
> $$
> When $\vec{F}$ is a fluid velocity field, the flux is the net volume of fluid
> crossing $S$ per unit time, positive in the direction of $\hat{n}$.

In parametric form the normal and the area element combine: since
$\hat{n}\,\d S = \dfrac{\vec{r}_u\times\vec{r}_v}{|\vec{r}_u\times\vec{r}_v|}\,|\vec{r}_u\times\vec{r}_v|\,\d A = (\vec{r}_u\times\vec{r}_v)\,\d A$,
the magnitude cancels and the flux is

$$
\iint\limits_S \vec{F}\cdot \d\vec{S} = \iint\limits_D \vec{F}(\vec{r}(u,v))\cdot(\vec{r}_u\times\vec{r}_v)\,\d A .
$$

Because of the cancellation, the square root in
$|\vec{r}_u\times\vec{r}_v|$ never appears in a flux computation. The
sign depends on the orientation, which fixes whether $\vec{r}_u\times\vec{r}_v$
or its negative is used.

$$
% caption: Flux counts the field's normal component crossing the surface; where
% $\mathbf{F}$ is perpendicular to $S$ the crossing is maximal, where tangent it
% is zero.
\begin{tikzpicture}[scale=1.0]
\definecolor{acc}{HTML}{4A6FA5}
% the surface patch
\fill[acc!8] (0.6,0.6) -- (3.6,0.6) -- (4.2,1.6) -- (1.2,1.6) -- cycle;
\draw[acc, thick] (0.6,0.6) -- (3.6,0.6) -- (4.2,1.6) -- (1.2,1.6) -- cycle;
\node[acc, font=\small, anchor=north east] at (3.6,0.6) {$S$};
% normal
\draw[black, thick, ->] (2.4,1.1) -- (2.15,2.1);
\node[font=\small, anchor=east] at (2.15,2.0) {$\mathbf{n}$};
% field arrows piercing the surface
\foreach \x in {1.3,2.4,3.4} {
  \draw[acc, thick, ->] (\x,-0.2) -- ({\x-0.15},1.4);
}
\node[acc, font=\small, anchor=north] at (1.15,-0.25) {$\mathbf{F}$};
\node[font=\small, anchor=west] at (3.6,1.9) {f\/lux through $S$};
\end{tikzpicture}
$$

> **Worked example.** Find the flux of $\vec{F} = \langle x, y, z\rangle$ outward
> through the unit sphere. With the outward parametrization, one finds
> $\vec{r}_\phi\times\vec{r}_\theta$ pointing outward with the field
> $\vec{F}(\vec{r}) = \vec{r}$ itself radial, so
> $\vec{F}\cdot(\vec{r}_\phi\times\vec{r}_\theta) = \sin\phi$ after
> simplification, and
>
> $$
> \iint\limits_S \vec{F}\cdot \d\vec{S} = \int_0^{2\pi}\int_0^\pi \sin\phi\,\d\phi\,\d\theta = 4\pi .
> $$

The result equals the volume of the ball times $3$; the
[Divergence Theorem](/calculus/multiple-integrals-and-vector-calculus/stokes-and-the-divergence-theorem)
will explain why, since $\operatorname{div}\vec{F} = 3$ and the ball has volume
$\tfrac{4}{3}\pi$.

## 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 | $1$ | $\iint\limits_D \abs{\vec{r}_u\times\vec{r}_v}\,\d A$ | area of $S$ |
| scalar surface integral | $f$ | $\iint\limits_D f(\vec{r})\,\abs{\vec{r}_u\times\vec{r}_v}\,\d A$ | mass of a shell, average value |
| flux | $\vec{F}\cdot\hat{n}$ | $\iint\limits_D \vec{F}(\vec{r})\cdot(\vec{r}_u\times\vec{r}_v)\,\d A$ | flow across $S$ |

Flux is the surface integral that appears in
[Stokes' Theorem and the Divergence Theorem](/calculus/multiple-integrals-and-vector-calculus/stokes-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.[^stewart166]

[^stewart166]: Stewart, §16.6 — Parametric Surfaces and Their Areas; §16.7 — Surface Integrals. The area element $|\vec{r}_u\times\vec{r}_v|\,\d A$, orientation by a continuous unit normal, and the flux integral $\iint\limits_S \vec{F}\cdot \d\vec{S} = \iint\limits_D \vec{F}\cdot(\vec{r}_u\times\vec{r}_v)\,\d A$.