Oriented Surface Flux

Flow through a surface

Flux counts the component of a vector field normal to an oriented surface:

\iint_S\mathbf F\cdot\mathbf n\,dS.

Positive flux goes in the selected normal direction. Negative flux crosses against it. A field tangent to the surface contributes zero, even when it is large. Reversing the normal reverses the answer.

Visual guide

VISUAL GUIDEOnly the normal component crosses the surface

Tangential part
Normal component

The shaded horizontal patch has upward normal n. The slanted vector F decomposes into a horizontal tangent component and an upward normal component. Flux uses the upward component times area; the tangent part contributes zero.

Parametrize and orient

Using $\mathbf r(u,v)$ , compute $\mathbf N=\mathbf r_u\times\mathbf r_v$ . Check its direction at an easy point. Negate it if necessary. Then integrate $\mathbf F(\mathbf r)\cdot\mathbf N\,du\,dv$ . Do not normalize $\mathbf N$ and forget the area factor: its magnitude already contains that factor.

Worked example: upward flow through a roof

The surface $z=x+y$ above $0\le x,y\le1$ has upward vector area $(-1,-1,1)dx\,dy$ . For $\mathbf F=(0,0,z)$ the dot product is $x+y$ , so upward flux is $1$ . The scalar surface integral of $z$ would be $\sqrt3$ , illustrating why flux and surface mass differ.

Worked example: radial flow through a sphere

On a sphere of radius $R$ , the outward unit normal is $\mathbf n=\mathbf r/R$ . For $\mathbf F=\mathbf r$ , $\mathbf F\cdot\mathbf n=R$ . Multiplying by area $4\pi R^2$ gives flux $4\pi R^3$ . The inward orientation yields its negative.

Orientation in practice

For a closed surface, outward orientation is conventional. An open surface needs an explicit choice such as upward, downward or away from an axis. Split a piecewise smooth surface into patches whose normals follow the same convention. The divergence and Stokes theorems in the following lessons offer alternative calculations when their conditions hold.

Practice

Find upward flux of $(0,0,3)$ through a horizontal disk of radius $2$ .
Find flux of $(1,0,0)$ through that disk.
Reverse the first disk’s orientation.

Show worked solutions

Normal component is $3$ and area is $4\pi$ , giving $12\pi$ .
The field is tangent, so the dot product is zero everywhere.
Downward flux is $-12\pi$ . The geometry is unchanged; only the selected direction changes.

Further study

MIT OpenCourseWare: Multivariable Calculus provides a full university course with additional lectures and exercises.