Flux

Section 11.1 Flux

At any give point along a curve, there is a natural vector, namely the (unit) tangent vector $\TT\text{.}$ Therefore, it is natural to add up the tangential component of a given vector field along a curve. When the vector field represents force, this integral represents the work done by the force along the curve. But there is no natural tangential direction at a point on a surface, or rather there are too many of them. The natural vector at a point on a surface is the (unit) normal vector $\nn\text{,}$ so on a surface it is natural to add up the normal component of a given vector field; this integral is known as the flux of the vector field through the surface.

We already know that the vector surface element is given by

\begin{equation} d\AA = d\rr_1 \times d\rr_2 .\tag{11.1.1} \end{equation}

Since $d\rr_1$ and $d\rr_2$ are both tangent to the surface, $d\AA$ is perpendicular to the surface, and is therefore often written

\begin{equation} d\AA = \nn \,\dA .\tag{11.1.2} \end{equation}

Putting this all together, the flux of a vector field $\FF$ through the surface is given by

\begin{equation} \hbox{flux of $\FF$ through $S$} = \Sint \FF\cdot d\AA .\tag{11.1.3} \end{equation}

Figure 11.1. A triangular region in the first octant, chopped parallel to the $x$- and $y$-axes.

We first consider a problem typical of those in calculus textbooks, namely finding the flux of the vector field $\FF=z\,\zhat$ up through the part of the plane $x+y+z=1$ lying in the first octant, as shown in Figure 11.1. We begin with the infinitesimal vector displacement in rectangular coordinates in 3 dimensions, namely

\begin{equation} d\rr = dx\,\xhat + dy\,\yhat + dz\,\zhat .\tag{11.1.4} \end{equation}

A natural choice of curves in this surface is given by setting $y$ or $x$ constant, so that $dy=0$ or $dx=0\text{,}$ respectively. We thus obtain

\begin{align*} d\rr_1 \amp = dx\,\xhat + dz\,\zhat = (\xhat-\zhat)\,dx ,\\ d\rr_2 \amp = dy\,\yhat + dz\,\zhat = (\yhat-\zhat)\,dy , \end{align*}

where we have used what we know (the equation of the plane) to determine each expression in terms of a single parameter. The surface element is thus

\begin{equation} d\AA = d\rr_1\times d\rr_2 = (\xhat+\yhat+\zhat)\,dx\,dy\tag{11.1.5} \end{equation}

and the flux becomes ¹

\begin{align} \Sint \FF\cdot d\AA \amp= \Sint z \> dx\,dy\notag\\ \amp= \int_0^1 \int_0^{1-y} (1-x-y) \,dx\,dy = \frac16 .\tag{11.1.6} \end{align}

The limits were chosen by visualizing the projection of the surface into the $xy$-plane, which is a triangle bounded by the $x$-axis, the $y$-axis, and the line whose equation is $x+y=1\text{.}$ Note that this latter equation is obtained from the equation of the surface by using what we know, namely that $z=0\text{.}$

Just as for line integrals, there is a rule of thumb which tells you when to stop using what you know to compute surface integrals: Don't start integrating until the integral is expressed in terms of two parameters, and the limits in terms of those parameters have been determined. Surfaces are two-dimensional!

Some readers will prefer to change the domain of integration in the second integral to be the projection of $S$ into the $xy$-plane. We prefer to integrate over the actual surface whenever possible.