Flux Integrals

Main idea

Flux measures how much of a flow passes through a surface, like how much air goes through a window screen each second. Only the part of the flow going straight through counts, so at each point you take the component of $\vec F$ along the normal and add it up over the surface.

\iint_S\vec F\cdot d\vec S = \iint_S\vec F\cdot{\htmlData{sym=0}{\vec n}}\,dS = \iint_D\vec F\big(\vec r(u, v)\big)\cdot\left({\htmlData{sym=1}{\vec r_u\times\vec r_v}}\right)du\,dv

What the symbols mean

$\vec n$
Unit normal
Chooses which way counts as positive: outward for closed surfaces. F·n is the part of the flow going straight through.
$\vec r_u\times\vec r_v$
Normal from the parametrization
Not a unit vector: its length already includes the area factor.

Explanation

Flux is simply the rate at which something flows through a surface: liters of water per second through a pipe's cross-section, or air per second through an open window. The notation $\iint_S\vec F\cdot\vec n\,dS$ is three familiar pieces: $\vec F$ (the flow at each point), $\vec n$ (a unit arrow sticking straight out of the surface) and $dS$ (a tiny patch of area). The dot product $\vec F\cdot\vec n$ keeps only the part of the flow going straight through; multiply by the patch's area and add up. That's all flux is.

Why the dot product. Picture holding a net in the wind. Face it straight into the wind and it catches the most air. Tilt it and less air gets through, and turn it edge-on and it catches nothing. Flow sliding along the surface doesn't cross it at all. The dot product with the unit normal measures exactly that: it's $\|\vec F\|\cos(\text{angle})$ , full strength head-on, zero edge-on. It's the same job the dot product did for work in line integrals, where only the part of the force along the path counted. There, it was the tangent direction; here, it's the normal direction.

Computing it. With a parametrization, you don't need to find the unit normal and $dS$ separately: $\vec n\,dS = (\vec r_u\times\vec r_v)\,du\,dv$ , because the cross product already points along the normal and its length is the area factor. So the flux is $\iint_D\vec F(\vec r(u, v))\cdot(\vec r_u\times\vec r_v)\,du\,dv$ .

Orientation. A surface has two sides, and flux depends on which normal you choose: flipping it flips the sign, like reversing a path for work. For closed surfaces (spheres, boxes), use the outward normal, so positive flux means net flow out.

A common mistake. It's tempting to take whatever $\vec r_u\times\vec r_v$ comes out and keep going. Always check which way it points. If the problem asks for upward or outward flux and your cross product points down or in, swap the order of the cross product (or flip the sign at the end).

Meaning. For a fluid velocity field, flux is volume per unit time passing through. For an electric field, it counts the field "lines" piercing the surface, which is what Gauss's law is about.

See it

Wind blowing straight up, F = ⟨0, 0, 2⟩, through a 1 × 1 net that tilts over. Notice the flux drops as it tilts: 2 when flat, about 1.41 at 45°, and 0 when it's stood on edge, since it's F · n times the area, 2 cos(tilt). The air through the tilted net is exactly the air through its shadow on the floor. Drag to turn it.

Worked examples

Example 1

Find the outward flux of $\vec F = \langle x, y, z\rangle$ through the sphere of radius 2.

1
On a sphere centered at the origin, the outward unit normal points along the position vector: $\vec n = \frac{\langle x, y, z\rangle}{2}$ . So $\vec F\cdot\vec n = \frac{x^2 + y^2 + z^2}{2} = \frac42 = 2$ at every point of the sphere.

Practice

Type your answer and check it. Solve them all to complete the lesson.

Problem 1+5 XP

Find the upward flux of $\vec F = \langle 0, 0, 1\rangle$ through the unit square in the $xy$ -plane.

Problem 2+5 XP

Find the outward flux of $\langle x, y, z\rangle$ through the sphere of radius 2.

Problem 3+5 XP

Find the outward flux of $\langle 0, 0, z\rangle$ through the upper unit hemisphere (not including the base).

Problem 4+5 XP

What is the flux of a field that is everywhere tangent to the surface?

Problem 5+5 XP

Find the upward flux of the constant field $\langle 1, 2, 3\rangle$ through the unit square in the plane $z = 0$ .

Problem 6+5 XP

Water flows through a pipe of radius 0.1 m with speed $v = 3\left(1 - \frac{r^2}{0.01}\right)$ m/s along the pipe: fastest at the center, zero at the wall. Find the flow rate (the flux through a cross-section), in liters per second.

Problem 7+5 XP

Find the upward flux of $\vec F = \langle x, y, z\rangle$ through the part of the plane $z = 2 - x - y$ in the first octant.

Challenge problem+25 XP

The hardest one here. Get it, and you've got this lesson.

Find the outward flux of $\vec F = \langle 0, 0, z\rangle$ through the whole closed cylinder $x^2 + y^2 \le 1$ , $0 \le z \le 2$ : the side, the top and the bottom.

Where it's used

Gauss's law says the electric flux out of any closed surface is proportional to the charge inside; engineers compute airflow through vents and heat flux through walls the same way.

How Much Smoke Leaves a RegionCalc III · Physics

Surface Integrals Stokes' Theorem