Suppose that \(D\) is a domain which can be described as
\begin{equation*}
D=\{(x_1,x_2)\in\mathbb R^2 \mid a\leq x_1\leq b,\varphi(x_1)\leq x_2\leq\psi(x_1)\},
\end{equation*}
where \(a<b\) are constants, and \(\varphi<\psi\) are continuously differentiable functions. Further assume that \(f_2=0\text{,}\) that is,
\begin{equation*}
\vect f=(f_1,0)\text{.}
\end{equation*}
To compute
\begin{equation*}
\int_{\partial D}f_1\,dx_1:=\int_{\partial D}f_1\,dx_1+0\,dx_2
\end{equation*}
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
\((b,t)\text{,}\) \(t\in[\varphi(b),\psi(b)]\text{.}\) According to the definition of the line integral (see
Definition 7.15) we get
\begin{equation*}
\int_{C_1}f_1\,dx_1
=\int_{\varphi(b)}^{\psi(b)}(f_1(b,t)\cdot 0+0\cdot 1)\,dt=0\text{.}
\end{equation*}
In much the same way we compute the integral over \(C_3\text{,}\) the reverse of which we can parametrise by \((a,t)\text{,}\) \(t\in[\varphi(a),\psi(a)]\text{.}\) As before we get
\begin{equation*}
\int_{C_3}f_1\,dx_1
=-\int_{\varphi(a)}^{\psi(a)}(f_1(a,t)\cdot 0+0\cdot 1)\,dt=0\text{.}
\end{equation*}
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
\begin{equation*}
\int_{C_2}f_1\,dx_1
=-\int_a^b\bigl(f_1(t,\psi(t))\cdot 1+0\cdot\psi'(t)\bigr)\,dt
=-\int_a^bf_1(t,\psi(t))\,dt\text{.}
\end{equation*}
In a similar manner we get the integral over \(C_4\text{.}\) Note however, that the orientation of the parametrisation \((t,\varphi(t))\text{,}\) \(t\in[a,b]\) is consistent with the orientation. Hence we get
\begin{equation*}
\int_{C_4}f_1\,dx_1
=\int_a^b\bigl(f_1(t,\varphi(t))\cdot 1+0\cdot\varphi'(t)\bigr)\,dt
=\int_a^bf_1(t,\varphi(t))\,dt\text{.}
\end{equation*}
Adding the four integrals together we finally get
\begin{equation}
\int_{\partial D}f_1\,dx_1
=-\int_a^b\bigl(f_1(t,\psi(t))-f_1(t,\varphi(t))\bigr)\,dt\text{.}\tag{10.2}
\end{equation}
We next want to rewrite the right hand side of the above equation. To do so we set for every fixed \(t\in[a,b]\)
\begin{equation*}
g(x_2):=f_1(t,x_2)
\end{equation*}
where \(g\) is defined for all \(x_2\in[\varphi(t),\psi(t)]\text{.}\) By the fundamental theorem of calculus we can write
\begin{equation*}
g(\psi(t))-g(\varphi(t)) =\int_{\varphi(t)}^{\psi(t)}\frac{d}{dx_2}g(x_2)\,dx_2\text{.}
\end{equation*}
By the definition of partial derivatives we have
\begin{equation*}
\frac{d}{dx_2}g(x_2)=\frac{\partial}{\partial x_2}f_1(t,x_2)\text{.}
\end{equation*}
Substituting all this into
(10.2) we get
\begin{equation*}
\int_{\partial D}f_1\,dx_1
=-\int_a^b\Bigl(\int_{\varphi(t)}^{\psi(t)}
\frac{\partial}{\partial x_2}f_1(t,x_2)\,dx_2\Bigr)\,dt\text{.}
\end{equation*}
Note that we can rename \(t\) by \(x_1\) (this was our choice for the parameter anyway). Then the formula becomes
\begin{equation*}
\int_{\partial D}f_1\,dx_1
=-\int_a^b\Bigl(\int_{\varphi(x_1)}^{\psi(x_1)}
\frac{\partial}{\partial x_2}f_1(x_1,x_2)\,dx_2\Bigr)\,dx_1
\end{equation*}
\begin{equation*}
\int_{\partial D}f_1\,dx_1
=-\int_D\frac{\partial}{\partial x_2}f_1(\vect x)\,d\vect x
\end{equation*}
