Vector Calculus: An Introduction

Suppose that $\mathit{f}=(f_1,f_2)$ is a smooth vector field. Applying Green’s theorem to the field $(-f_2,f_1)$ we obtain

We want to get an interpretation of the integral on the right hand side. To do so suppose that

is a regular parametrisation of some part $C\subset \partial D$ consistent with the orientation of $\partial D\text{.}$ By definition of the line integral we have

We can view the integrand on the right hand side as a dot product of $\mathit{f}$ with the vector $({\gamma }_2^{\prime }(t),-{\gamma }_1^{\prime }(t))\text{.}$ We rescale the latter to unit length and define

Observe that $\mathit{n}(t)$ is perpendicular to the positive unit tangent vector $\mathit{\tau }(t)={\mathit{\gamma }}^{\prime }(t)/\Vert {\gamma }^{\prime }(t)\Vert \text{.}$ As it is outward pointing, it is called the outward pointing unit normal to $D\text{.}$ Both vectors are shown in Figure 11.1.

Hence we can rewrite our integral as

By our Definition 7.15 of line integrals we see that

Let $\mathrm{\Delta }s$ denote a very small line element. If $\mathit{f}$ models the motion of a fluid (direction and velocity of fluid at a given point) then $\mathit{f}\cdot \mathit{n}\mathrm{\Delta }s$ is the flux across $\partial D$ through the line element $\mathrm{\Delta }s\text{,}$ and $\mathit{f}\cdot \mathit{n}\mathrm{\Delta }s$ is the approximate amount of fluid flowing across the boundary $\partial D$ through $ds\text{.}$ The situation is depicted in Figure 11.2. Note that the area of the parallelogram and the rectangle are the same.

Hence the integral on the right hand side of (11.1) represents the total flux of $\mathit{f}$ across $C\subset \partial D\text{.}$ Note that the above arguments are completely analogous to those used in Section 9.6 to find the flux of a vector field in space across a surface. Going back to the original formula we see that

is the total loss (or increase) of the fluid from $D$ through $\partial D\text{.}$ We make the following definition.