Vector Calculus: An Introduction

Suppose that $D$ is a domain which can be described as

where $a<b$ are constants, and $\phi <\psi$ are continuously differentiable functions. Further assume that $f_2=0\text{,}$ that is,

To compute

we determine the path integral over the curves $C_1$-$C_4$ indicated in Figure 10.4, and then add them up to get the total integral.

We first look at $C_1\text{,}$ which we can parametrise by $(b,t)\text{,}$ $t\in [\phi (b),\psi (b)]\text{.}$ According to the definition of the line integral (see Definition 7.15) we get

In much the same way we compute the integral over $C_3\text{,}$ the reverse of which we can parametrise by $(a,t)\text{,}$ $t\in [\phi (a),\psi (a)]\text{.}$ As before we get

We next compute the integral over $C_2\text{.}$ We can parametrise the reverse of $C_2$ by $(t,\psi (t))\text{,}$ $t\in [a,b]\text{.}$ Note that this parametrisation is against the orientation of $C_2\text{,}$ so we get

In a similar manner we get the integral over $C_4\text{.}$ Note however, that the orientation of the parametrisation $(t,\phi (t))\text{,}$ $t\in [a,b]$ is consistent with the orientation. Hence we get

Adding the four integrals together we finally get

We next want to rewrite the right hand side of the above equation. To do so we set for every fixed $t\in [a,b]$

where $g$ is defined for all $x_2\in [\phi (t),\psi (t)]\text{.}$ By the fundamental theorem of calculus we can write

By the definition of partial derivatives we have

Substituting all this into (10.2) we get

Note that we can rename $t$ by $x_1$ (this was our choice for the parameter anyway). Then the formula becomes

Applying Fubini’s Theorem 5.12 we finally get

We next consider the special case with $D$ of the form

where $a<b$ are constants and $\phi <\psi$ two continuously differentiable functions on $[a,b]\text{.}$ Moreover, let us assume that $f_1=0\text{,}$ so that $\mathit{f}=(0,f_2)\text{.}$ We can reduce this case to the previous one by interchanging the role of $x_1$ and $x_2\text{.}$

Note however, that this changes the orientation of the boundary of $D\text{,}$ changing the sign in the line integral. Hence

Let us now assume that $D$ admits a representation of the form considered in Case 1 and Case 2 simultaneously. This is the case if every horizontal and vertical line intersect the domain in exactly two points or exactly one line segment. First we use the fact that $D$ admits a representation considered in Case 1. Applying this to the vector field $(f_1,0)$ we get

Next we use the fact that $D$ admits a representation considered in Case 2. Applying this to the vector field $(0,f_2)$ we get

Adding these formluas we get equation (10.1). This proves Green’s theorem for the special domains considered.