Green's theorem

We get some definitions out of the way first, before we state the theorem. We start with a closed, simple curve $\mathcal{C}$ which encloses a region $\mathcal{R}$ (see Fig. 2.10). A simple curve is one that does not cross itself. Since the curve is closed and simple, the enclosed region $\mathcal{R}$ does not have any holes. In Subsec. 2.3.2, we discussed the orientation of some curve, $\mathcal{C}$ (which was not necessarily closed). In the case of closed curves, we use the convention that a simple, closed curve has positive orientation if it is traced out in a counter-clockwise direction (as indicated by the arrows in Fig. 2.10). We take a curve that is traced out clockwise to indicate a negative orientation. Another way of thinking about this is the following: imagine you are a point moving along the closed curve in the positive orientation. Then, the enclosed region $\mathcal{R}$ is always on your left.

Figure 2.10: A simple, closed curve $\mathcal{C}$, traced out in the counter-clockwise direction, indicating a positive orientation. The curve encloses a region $\mathcal{R}$.

Green's theorem

For any region $\mathcal{R}$ enclosed by a boundary $\mathcal{C}$ representing a positively oriented, piecewise smooth, simple, closed curve and a vector field $\boldsymbol{F}=$ $\left(F_{1}, F_{2}\right)$, where $F_{1}, F_{2}$ are continuous with continuous partial derivatives, then:

The notation $\oint_{\mathcal{C}} F_{1} d x+F_{2} d y$ assumes that the curve $\mathcal{C}$ satisfies the conditions given by Green's theorem.

Remarks:

• If $\boldsymbol{F}$ is conservative, then $\frac{\partial F_{1}}{\partial y}=\frac{\partial F_{2}}{\partial x}$ and so $\oint_{\mathcal{C}} F_{1} d x+F_{2} d y=0$ on a closed path;
• Green's theorem can be thought of as generalisations to the 'Fundamental Theorem of Calculus' for double integrals. The former shows that integration and differentiation are inverse operations to each other, i.e. $G(b)-G(a)=\int_{a}^{b} G^{\prime}(x) d x$
• The theorem relates the behavior of $\boldsymbol{F}$ at the boundary of $\boldsymbol{F}$ with the behavior of its derivatives inside $\boldsymbol{F}$.

Next, we look at an example where we use Green's theorem to evaluate a line integral.

Figure 2.11: The curve $\mathcal{C}$ given by the right angled triangle whose vertices are indicated in Example 2.15. Note the positive orientation is counter-clockwise, as shown by the direction of the arrows. Example 2.15 Use Green's theorem to evaluate

where $\mathcal{C}$ is the triangle with vertices $(0,0),(1,0),(1,2)$ with positive orientation.

Solution Figure 2.11 shows the curve $\mathcal{C}$ given by a right angled triangle. Before we use Green's theorem, we want to ensure that the conditions specified in Theorem 2.10 are met. The curve is simple, closed and positively oriented and the enclosed region $\mathcal{R}$ is defined by $0 \leq x \leq 1$ and $0 \leq y \leq 2 x$. From the problem statement, we identify $F_{1}=x y$ and $F_{2}=x^{2} y^{3}$. We compute the partial derivatives:

Using Eq. (2.112), we have:

Regions with holes

Green's theorem as stated does not apply with regions with holes in them (since we need a simply connected region to apply the theorem). However, the theorem can be extended to regions with holes such that it is bounded by two piecewise smooth, simple, closed curves. Figure 2.12(a) shows a region $\mathcal{R}$ bounded by two curves $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$ (shaded region). The union of the two curves describes the boundary of $\mathcal{R}$, i.e. $\mathcal{C}=\mathcal{C}_{1} \cup \mathcal{C}_{2} 1$. Notice that the curve $\mathcal{C}$ is positively oriented since if we traverse the outer curve in the counterclockwise direction and the inner curve in the clockwise direction [as shown in Fig. 2.12(a)], the region $\mathcal{R}$ is always to the left. To apply Green's theorem, we cut the region in half (such that we have $\mathcal{R}=\mathcal{R}_{1} \cup \mathcal{R}_{2}$ ) and rename the various portions of the curves. This is shown in Fig. 2.12(b). The top and bottom parts of the original curve $\mathcal{C}_{1}$ are now called $\mathcal{C}_{1_{\mathrm{t}}}$ and $\mathcal{C}_{1_{\mathrm{b}}}$, respectively. Similarly, we have $\mathcal{C}_{2 \mathrm{t}}$ and $\mathcal{C}_{2 \mathrm{~b}}$ for the top and bottom parts of the curve $\mathcal{C}_{2}$. Next, the crosscuts (that divide the region into $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$ ) are called $\mathcal{C}_{3}$ and $\mathcal{C}_{4}$; note that the opposite orientation of the curves includes a negative sign. We can now describe the boundaries of the two subregions through the different curves. We have the boundary of $\mathcal{R}_{1}$ given by a curve $\mathcal{U}_{1}$ where $\mathcal{U}_{1}=\mathcal{C}_{1_{\mathrm{t}}} \cup \mathcal{C}_{2_{\mathrm{t}}} \cup \mathcal{C}_{3} \cup \mathcal{C}_{4}$ and the boundary of $\mathcal{R}_{2}$ is $\mathcal{U}_{2}$, where $\mathcal{U}_{2}=\mathcal{C}_{1 \mathrm{~b}} \cup \mathcal{C}_{2 \mathrm{~b}} \cup\left(-\mathcal{C}_{3}\right) \cup\left(-\mathcal{C}_{4}\right)$. Notice that we can now use Green's theorem on each of these subregions since they do not have holes.

(a)

(b)

Figure 2.12: (a) The shaded region $\mathcal{R}$ bounded by two curves, the outher curve $\mathcal{C}_{1}$ and the inner curve, $\mathcal{C}_{2}$. (b) The region $\mathcal{R}$ divided into 2 subregions $\mathcal{R}_{1}$ and $\mathcal{R}_{2}$. The crosscuts given by the curves $\mathcal{C}_{3}$ and $\mathcal{C}_{4}$ are added to describe the boundaries of the two subregions.

${ }^{1}$ The symbol $\cup$ represents the union which means that $\mathcal{C}$ consists of both $\mathcal{C}_{1}$ and $\mathcal{C}_{2}$. Using the result from Theorem 10, we have the following integrals over $d A=d x d y$ :

where $\mathcal{U}_{1}$ and $\mathcal{U}_{2}$ are given by the union of the curves forming the boundaries of the two subregions as given above. Now, we can break the line integrals on each piece of the boundary. We break the line integral over $\mathcal{U}_{1}$ as follows:

Similarly, we break the line integral over $\mathcal{U}_{2}$ :

Adding Eqs. (2.114) and (2.115), results in the curves that have the same boundary but opposite orientation (i.e. the line integrals over $\mathcal{C}_{3},-\mathcal{C}_{3}$ and $\mathcal{C}_{4}$, $-\mathcal{C}_{4}$ ) cancelling each other. And hence,

Notice that we can now add up the line integrals to make $\mathcal{C}=\mathcal{C}_{1_{\mathrm{t}}} \cup \mathcal{C}_{2_{\mathrm{t}}} \cup \mathcal{C}_{1_{\mathrm{b}}} \cup \mathcal{C}_{2_{\mathrm{b}}}$, such that:

which is the result from Green's theorem. Therefore for the shape shown in Fig. 2.12. we could use Green's theorem as stated to evaluate $\mathcal{R}$. This is generally true for regions that have holes in them.