Scalar Surface Integrals and Flux, multivariable and vector calculus lesson

A thin metal dome is thicker near the top than near the rim, and you want its mass. Or wind blows through a fishing net, and you want to know how much air passes through it each second. Both questions add up small pieces of a curved surface. The first needs only the area of each piece. The second also needs to know which way each piece faces, and which side counts as “through”.

Key terms

Surface element

Definition Let $D$ be a closed bounded region in the $uv$ -plane with piecewise smooth boundary, and let $\mathbf{r}(u, v)$ be a regular parametrization of a surface $S$ , meaning $\mathbf{r}$ has continuous first partial derivatives, is one-to-one on the interior of $D$ , and has $\mathbf{r}_u \times \mathbf{r}_v \ne \mathbf{0}$ there. The surface element is

dS = \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, du \, dv.

For a graph $z = f(x, y)$ over $D$ , with $f$ having continuous first partial derivatives, this becomes

dS = \sqrt{1 + f_x^2 + f_y^2} \, dA.

In plain words A small rectangle of size $du$ by $dv$ in the parameter plane is stretched onto the surface as a small, nearly flat parallelogram. The length of the cross product is the factor that turns parameter area into actual area.

Example On the sphere of radius $a$ use

\begin{aligned} \mathbf{r}(\phi, \theta) = \langle &a \sin\phi \cos\theta, \\ &a \sin\phi \sin\theta,\ a \cos\phi \rangle, \end{aligned}

with $0 \le \phi \le \pi$ and $0 \le \theta \le 2\pi$ . A direct computation gives $\mathbf{r}_\phi \times \mathbf{r}_\theta = a \sin\phi \, \mathbf{r}$ , so

dS = a^2 \sin\phi \, d\phi \, d\theta.

The factor $\sin\phi$ is never negative on $[0, \pi]$ . It vanishes only at the poles, two points that carry no area.

Surface integral

Definition Let $S$ have a regular parametrization $\mathbf{r}$ on $D$ as above, and let $g$ be continuous on $S$ . The surface integral of $g$ over $S$ is

\begin{aligned} &\iint_S g \, dS \\[0.4em] &= \iint_D g(\mathbf{r}(u, v)) \, \lvert \mathbf{r}_u \times \mathbf{r}_v \rvert \, du \, dv. \end{aligned}

Its value does not depend on which regular parametrization of $S$ you use. For a piecewise smooth surface, add the integrals over the smooth pieces. With $g = 1$ you get the surface area of $S$ .

In plain words Evaluate $g$ at a point of the surface, multiply by the actual area of the little patch there, and add. The parametrization only tells you how to reach the points and how much area each parameter step covers.

Example Take the upper unit hemisphere, $x^2 + y^2 + z^2 = 1$ with $z \ge 0$ , and $g = z$ . In spherical parameters $g = \cos\phi$ and $dS = \sin\phi \, d\phi \, d\theta$ , so

\begin{aligned} &\iint_S z \, dS \\[0.4em] &= \int_0^{2\pi} \!\! \int_0^{\pi/2} \cos\phi \sin\phi \, d\phi \, d\theta \\[0.4em] &= 2\pi \cdot \frac{1}{2} = \pi. \end{aligned}

Density and mass of a surface

Definition A thin sheet modeled by a surface $S$ has area density $\delta(x, y, z)$ , a continuous nonnegative function measured in mass per unit area. Its mass is

m = \iint_S \delta \, dS.

In plain words Density tells you how heavy each square centimeter is at that spot. Multiply by the area of each patch and add.

Example A lampshade has the shape of the cone $z = \sqrt{x^2 + y^2}$ for $0 \le z \le 1$ , measured in centimeters, with density $\delta = z$ grams per square centimeter. The graph formula needs care, because $f(x, y) = \sqrt{x^2 + y^2}$ has no partial derivatives at the origin. Use the parametrization

\mathbf{r}(r, \theta) = \langle r \cos\theta,\ r \sin\theta,\ r \rangle

on the rectangle $0 \le r \le 1$ , $0 \le \theta \le 2\pi$ instead. It has continuous partial derivatives on the closed rectangle, and

\mathbf{r}_r \times \mathbf{r}_\theta = \langle -r \cos\theta, -r \sin\theta, r \rangle

has length $\sqrt{2}\, r$ . That is nonzero, and $\mathbf{r}$ is one-to-one, on the interior $0 < r < 1$ , $0 < \theta < 2\pi$ , so this is a regular parametrization in the sense above. Its only failure is on the edge $r = 0$ , a segment of zero area in the parameter plane that is squeezed to the vertex. So $dS = \sqrt{2}\, r \, dr \, d\theta$ and $z = r$ , which gives

\begin{aligned} m &= \int_0^{2\pi} \int_0^1 r \cdot \sqrt{2} \, r \, dr \, d\theta \\[0.4em] &= \frac{2\sqrt{2}\,\pi}{3} \approx 2.96 \text{ grams.} \end{aligned}

Equivalently, integrate over $\varepsilon \le r \le 1$ , where the cone is a smooth graph, and let $\varepsilon \to 0$ . The integrand is bounded, so the missing piece contributes at most a constant times $\varepsilon^2$ . Keep the three factors apart when you set this up. The density $r$ is in grams per square centimeter, the area factor $\sqrt{2}\, r \, dr \, d\theta$ is in square centimeters, and their product is in grams.

Oriented surface

Definition A surface $S$ is orientable if it has a unit normal vector $\mathbf{n}$ at every point off its edges and corners that varies continuously over each smooth piece and fits together consistently across the edges. A choice of such a normal field is an orientation, and $S$ together with it is an oriented surface. A connected orientable surface has exactly two orientations, $\mathbf{n}$ and $-\mathbf{n}$ .

In plain words Orienting a surface means picking one of its two sides and sticking with it everywhere, with no sudden flips.

Example The graph $z = 4 - x^2 - y^2$ over the disk $x^2 + y^2 \le 4$ can be oriented upward, with every normal having positive $z$ -component, or downward. A normal that points up at the top and down near the rim would not be continuous, so it is not an orientation.

Unit normal

Definition For a regular parametrization the unit normal it induces is

\mathbf{n} = \frac{\mathbf{r}_u \times \mathbf{r}_v}{\lvert \mathbf{r}_u \times \mathbf{r}_v \rvert}.

For a graph $z = f(x, y)$ , parametrized by $\langle x, y, f(x, y) \rangle$ , this gives the upward unit normal

\mathbf{n} = \frac{\langle -f_x, -f_y, 1 \rangle}{\sqrt{1 + f_x^2 + f_y^2}}.

Swapping the roles of $u$ and $v$ flips the sign, since $\mathbf{r}_v \times \mathbf{r}_u = -\mathbf{r}_u \times \mathbf{r}_v$ .

In plain words The cross product of the two edge vectors of a patch sticks straight out of the surface. Dividing by its length leaves a direction only.

Example On the sphere of radius $a$ the normal from the parametrization above is

\frac{a \sin\phi \, \mathbf{r}}{a^2 \sin\phi} = \frac{\mathbf{r}}{a} = \frac{\langle x, y, z \rangle}{a},

which points away from the center. Always check the direction geometrically. Near the north pole, where $\phi$ is small, this normal points up, as an outward normal should.

Vector flux

Definition Let $S$ be an oriented piecewise smooth surface with unit normal $\mathbf{n}$ , and let $\mathbf{F}$ be a vector field continuous on $S$ . The flux of $\mathbf{F}$ across $S$ is

\iint_S \mathbf{F} \cdot d\mathbf{S} = \iint_S \mathbf{F} \cdot \mathbf{n} \, dS.

If $\mathbf{r}$ is a regular parametrization on $D$ whose normal $\mathbf{r}_u \times \mathbf{r}_v$ points the same way as $\mathbf{n}$ , the lengths cancel and

\begin{aligned} &\iint_S \mathbf{F} \cdot d\mathbf{S} \\[0.4em] &= \iint_D \mathbf{F}(\mathbf{r}) \cdot (\mathbf{r}_u \times \mathbf{r}_v) \, du \, dv. \end{aligned}

In plain words Only the part of the field that crosses the surface counts, and it counts positive when it crosses toward the chosen side. If $\mathbf{F}$ is the velocity of a fluid, the flux is the volume crossing $S$ per unit time.

Example Find the flux of $\mathbf{F} = \langle y, x, z \rangle$ upward through the part of $z = 4 - x^2 - y^2$ with $z \ge 0$ . The upward normal vector is $\langle 2x, 2y, 1 \rangle$ , which has positive $z$ -component as required. On the surface

\begin{aligned} &\mathbf{F} \cdot \langle 2x, 2y, 1 \rangle \\[0.4em] &= 2xy + 2xy + (4 - x^2 - y^2) \\[0.4em] &= 4xy + 4 - r^2. \end{aligned}

The term $4xy$ integrates to $0$ over the disk of radius $2$ by symmetry, so

\begin{aligned} \int_0^{2\pi} \!\! \int_0^2 (4 - r^2)\, r \, dr \, d\theta &= 2\pi (8 - 4) \\[0.4em] &= 8\pi. \end{aligned}

Orientation reversal

Definition If $S$ is oriented by $\mathbf{n}$ and $-S$ denotes the same surface oriented by $-\mathbf{n}$ , then

\iint_{-S} \mathbf{F} \cdot d\mathbf{S} = -\iint_S \mathbf{F} \cdot d\mathbf{S},

while a scalar surface integral $\iint_S g \, dS$ is the same for both, since $dS$ is an area and is never negative.

In plain words Flux keeps track of direction, so choosing the other side flips its sign. Mass and area do not care which side you look from.

Example For the paraboloid above with the downward normal, the flux of $\langle y, x, z \rangle$ is $-8\pi$ . If the parametrization you picked gives the wrong normal, you may keep it and simply change the sign of the result, or swap the order of the cross product.

Nonorientable surfaces

Definition A surface is nonorientable if no continuous unit normal field exists on it. The standard example is the Möbius band. Scalar surface integrals are still defined on such a surface, but flux is not, because there is no consistent side to count as positive.

In plain words On a Möbius band you can walk once around the center line and come back on the other side of the paper without crossing an edge, so “this side” has no meaning.

Example Parametrize a Möbius band by

\begin{aligned} \mathbf{r}(u, v) = \big\langle &(2 + v \cos\tfrac{u}{2}) \cos u, \\ &(2 + v \cos\tfrac{u}{2}) \sin u, \\ &v \sin\tfrac{u}{2} \big\rangle, \end{aligned}

with $0 \le u \le 2\pi$ and $-1 \le v \le 1$ . Along the center line $v = 0$ a computation gives

\begin{aligned} \mathbf{r}_u \times \mathbf{r}_v = \big\langle &2 \cos u \sin\tfrac{u}{2}, \\ &2 \sin u \sin\tfrac{u}{2},\ -2 \cos\tfrac{u}{2} \big\rangle. \end{aligned}

At $u = 0$ this is $\langle 0, 0, -2 \rangle$ , and at $u = 2\pi$ , the same point $(2, 0, 0)$ , it is $\langle 0, 0, 2 \rangle$ . Following the normal continuously once around brings it back reversed.

Outward normal on a closed surface

Definition A closed surface is the boundary of a bounded solid region $E$ , for example a sphere or the surface of a box. Unless stated otherwise it is oriented by the outward unit normal, which points away from $E$ . If the surface is piecewise smooth, the flux is the sum of the fluxes over the smooth pieces, each with its own outward normal.

In plain words On a closed surface, positive flux means a net flow out of the solid. Each face gets the normal that leaves the solid there, so faces on opposite sides get opposite normals.

Example Let $E$ be the solid cylinder $x^2 + y^2 \le 1$ , $0 \le z \le 2$ , and $\mathbf{F} = \langle x, y, z \rangle$ .

Piece	Outward $\mathbf{n}$	$\mathbf{F} \cdot \mathbf{n}$	Area	Flux
Side	$\langle x, y, 0 \rangle$	$1$	$4\pi$	$4\pi$
Top $z = 2$	$\langle 0, 0, 1 \rangle$	$2$	$\pi$	$2\pi$
Bottom $z = 0$	$\langle 0, 0, -1 \rangle$	$0$	$\pi$	$0$

The outward flux is $6\pi$ . On the bottom the outward normal points down, not up, which is the most common sign error with piecewise closed surfaces.

Check yourself

Find the integral of $x^2 + y^2$ over the sphere $x^2 + y^2 + z^2 = 4$ .

Show the answer

By symmetry the integrals of $x^2$ , $y^2$ and $z^2$ over the sphere are equal, and their sum is the integral of $4$ , which is $4 \cdot 16\pi = 64\pi$ . So

\begin{aligned} \iint_S (x^2 + y^2) \, dS &= \frac{2}{3} \cdot 64\pi \\[0.4em] &= \frac{128\pi}{3}. \end{aligned}

Find the flux of $\mathbf{F} = \langle 0, 0, z \rangle$ upward through the part of the plane $x + y + z = 1$ in the first octant.

Show the answer

Write the plane as $z = 1 - x - y$ over the triangle $T$ with $x, y \ge 0$ and $x + y \le 1$ . The upward normal vector is $\langle 1, 1, 1 \rangle$ , so

\begin{aligned} &\iint_T (1 - x - y) \, dA \\[0.4em] &= \int_0^1 \!\! \int_0^{1 - x} (1 - x - y) \, dy \, dx = \frac{1}{6}. \end{aligned}

Let $S$ be the upper unit hemisphere oriented toward the origin. What is the flux of $\mathbf{F} = \langle 0, 0, 1 \rangle$ across $S$ ?

Show the answer

With the outward normal $\langle x, y, z \rangle$ the integrand is $\mathbf{F} \cdot \mathbf{n} = z$ , and the integral of $z$ over the hemisphere is $\pi$ . Pointing toward the origin reverses the orientation, so the flux is $-\pi$ .

Scalar Surface Integrals and Flux

Key terms

Surface element

Surface integral

Density and mass of a surface

Oriented surface

Unit normal

Vector flux

Orientation reversal

Nonorientable surfaces

Outward normal on a closed surface

Check yourself

How this lesson connects