Flux

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Section 12.3 Flux

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. 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{12.3.1} \end{equation}

A natural choice of curves in this surface is given by setting \(y\) or \(x\) constant, so that \(dy=0\) or \(dx=0\text{:}\)

\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{12.3.2} \end{equation}

and the flux becomes

¹

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.

\begin{equation} \Sint \FF\cdot d\AA = \Sint z \> dA = \int_0^1 \int_0^{1-y} (1-x-y) \,dx\,dy = \frac16 .\tag{12.3.3} \end{equation}

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!

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 \(z=0\text{.}\)