Setting Things Up

Suppose that $\vec{F} = F_1(x,y)\hat{i} + F_2(x,y)\hat{j}$ and C is a closed curve in the xy plane, oriented counter-clockwise as shown.

To calculate $\oint_C \vec{F} \cdot d\vec{r}$, break the area inside C (let's call this area R) up into smaller curves, $\Delta C_i$. Make sure that each of the $\Delta C_i$'s are oriented counter-clockwise as well. Notice that we can now write the circulation of $\vec{F}$ as

$$\oint_C \vec{F} \cdot d\vec{r} = \sum_i \oint_{\Delta C_i} \vec{F} \cdot d\vec{r}.$$ (4)

The reason this works is shown below.

Look at the boundary of two of thess curves. Since the line integral along one of these curves is the opposite of the line integral along the other (they're following the boundary between the curves in the opposite directions) the total constribution from this inner boundary is zero. Similarly, all of the other inner boundaries cancel, leaving only the contribution from the outside boundary, which is simply C itself.

Now we look at $\oint_{\Delta C_i}\vec{F} \cdot d\vec{r}$. To calculate this, let's assume that the dimensions of the curve are small enough that a linear approximation to $\vec{F}$ is good enough. We will treat the lower left corner of the box-shaped curve as having coordinates (x_i, y_i) and let the box have dimensions $\Delta x$ and $\Delta y$.