``The Proof''

Our linear approximation for $\vec{F}$ will look like

$$\vec{F}(x,y) = F_1 \hat{i} + F_2 \hat{j} = (a + b(x-x_i) + c(y - y_i))\hat{i} + (m + n(x - x_i) + p(y-y_i))\hat{j},$$ (5)

where we have made the following substitutions:

$$a = F_1(x_i,y_i) \qquad m = F_2(x_i,y_i)$$

$$b = \frac{\partial F_1}{\partial x}(x_i,y_i) \qquad n = \frac{\partial F_2}{\partial x}(x_i,y_i)$$

$$c = \frac{\partial F_1}{\partial y}(x_i,y_i) \qquad p = \frac{\partial F_2}{\partial y}(x_i,y_i).$$

Now $\oint_{\Delta C_i} \vec{F} \cdot d\vec{r} = \oint_{\Delta C_i}(F_1 dx + F_2 dy)$ which can be written as

$$\oint_{\Delta C_i} (a + b(x-x_i) + c(y-y_i))dx + (m + n(x-x_i) + p(y-y_i))dy.$$ (6)

It should be easy to convince yourself that the parts of the integral involving b(x-x_i)dx and p(y-y_i)dy will cancel in the end, since these represent pieces of a conservative vector field, whose circulation is always zero. Likewise, a constant vector field (the pieces adx and mdy) are also conservative. The line integral now reduces to

$$\oint_{\Delta C_i} \vec{F} \cdot d\vec{r} \approx \oint_{\Delta C_i} c(y - y_i)dx + \oint_{\Delta C_i} n(x - x_i)dy$$

$$= \oint_{C_1} cy dx + \oint_{C_3} cy dx + \oint_{C_2} nx dy + \oint_{C_4} nx dy$$

$$= cy_i \Delta x - c(y_i + \Delta y)\Delta x + n(x_i + \Delta x)\Delta y - nx_i \Delta y$$

$$= -c \Delta y \Delta x + n \Delta x \Delta y$$

$$= \left( -\frac{\partial F_1}{\partial y} - \frac{\partial F_2}{\partial x} \right) \Delta x \Delta y.$$

Now we can calculate the total circulation.

$$\oint_C \vec{F} \cdot d\vec{r} = \lim_{\vert\vert\Delta C_i\... ... 0} \sum_{\Delta C_i} \oint_{\Delta C_i} \vec{F} \cdot d\vec{r}$$

$$= \lim_{\vert\vert\Delta C_i\vert\vert\rightarrow 0} \sum_{\... ... y} - \frac{\partial F_2}{\partial x} \right) \Delta x \Delta y$$

$$= \int \int_R \left( -\frac{\partial F_1}{\partial y} - \frac{\partial F_2}{\partial x} \right) dx dy.$$