Green’s Theorem on General Regions

Green’s theorem, as stated, applies only to regions that are simply connected—that is, Green’s theorem cannot handle regions with holes. Here, we extend Green’s theorem so that it does work on regions with finitely many holes (Figure 1).

Before discussing extensions of Green’s theorem, we need to go over some terminology regarding the boundary of a region. Let [latex]D[/latex] be a region and let [latex]C[/latex] be a component of the boundary of [latex]D[/latex].

• We say that [latex]C[/latex] is positively oriented if, as we walk along [latex]C[/latex] in the direction of orientation, region [latex]D[/latex] is always on our left. The counterclockwise orientation of the boundary of a disk is a positive orientation, for example.
• Curve [latex]C[/latex] is negatively oriented if, as we walk along [latex]C[/latex] in the direction of orientation, region [latex]D[/latex] is always on our right. The clockwise orientation of the boundary of a disk is a negative orientation.

Let [latex]D[/latex] be a region with finitely many holes (so that [latex]D[/latex] has finitely many boundary curves), and denote the boundary of [latex]D[/latex] by [latex]\partial D[/latex] (Figure 2). To extend Green’s theorem so it can handle [latex]D[/latex], we divide region [latex]D[/latex] into two regions, [latex]D_1[/latex] and [latex]D_2[/latex] (with respective boundaries [latex]\partial D_1[/latex] and [latex]\partial D_2[/latex]), in such a way that [latex]D=D_1\cup D_2[/latex] and neither [latex]D_1[/latex] nor [latex]D_2[/latex] has any holes (Figure 2).

Assume the boundary of [latex]D[/latex] is oriented as in the figure, with the inner holes given a negative orientation and the outer boundary given a positive orientation. The boundary of each simply connected region [latex]D_1[/latex] and [latex]D_2[/latex] is positively oriented. If [latex]\mathbf{F}[/latex] is a vector field defined on [latex]D[/latex], then Green’s theorem says that:

[latex]\begin{aligned} \displaystyle\oint_{\partial{D}}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\oint_{\partial{D_1}}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\oint_{\partial{D_2}}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\displaystyle\iint_{D_1}Q_x-P_ydA+\displaystyle\iint_{D_2}Q_x-P_ydA \\ &=\displaystyle\iint_{D}(Q_x-P_y)dA \end{aligned}[/latex].

Therefore, Green’s theorem still works on a region with holes.

To see how this works in practice, consider annulus [latex]D[/latex] in Figure 3 and suppose that [latex]\mathbf{F}=\langle P,Q\rangle[/latex] is a vector field defined on this annulus. Region [latex]D[/latex] has a hole, so it is not simply connected.

Orient the outer circle of the annulus counterclockwise and the inner circle clockwise (Figure 3) so that, when we divide the region into [latex]D_1[/latex] and [latex]D_2[/latex], we are able to keep the region on our left as we walk along a path that traverses the boundary. Let [latex]D_1[/latex] be the upper half of the annulus and [latex]D_2[/latex] be the lower half. Neither of these regions has holes, so we have divided [latex]D[/latex] into two simply connected regions.

We label each piece of these new boundaries as [latex]P_i[/latex] for some [latex]i[/latex], as in Figure 3. If we begin at [latex]P[/latex] and travel along the oriented boundary, the first segment is [latex]P_1[/latex], then [latex]P_2[/latex], [latex]P_3[/latex], and [latex]P_4[/latex]. Now we have traversed [latex]D_1[/latex] and returned to [latex]P[/latex].

Next, we start at [latex]P[/latex] again and traverse [latex]D_2[/latex]. Since the first piece of the boundary is the same as [latex]P_4[/latex] in [latex]D_1[/latex], but oriented in the opposite direction, the first piece of [latex]D_2[/latex] is [latex]-P_4[/latex]. Next, we have [latex]P_5[/latex], then [latex]-P_2[/latex], and finally [latex]P_6[/latex].

Figure 3 shows a path that traverses the boundary of [latex]D[/latex]. Notice that this path traverses the boundary of region [latex]D_1[/latex], returns to the starting point, and then traverses the boundary of region [latex]D_2[/latex]. Furthermore, as we walk along the path, the region is always on our left.

Notice that this traversal of the [latex]P_i[/latex] paths covers the entire boundary of region [latex]D[/latex]. If we had only traversed one portion of the boundary of [latex]D[/latex], then we cannot apply Green’s theorem to [latex]D[/latex].

The boundary of the upper half of the annulus, therefore, is [latex]P_1\cup P_2\cup P_3\cup P_4[/latex] and the boundary of the lower half of the annulus is [latex]-P_4\cup P_5\cup -P_2\cup P_6[/latex]. Then, Green’s theorem implies:

[latex]\begin{aligned} \displaystyle\int_{\partial{D}}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int+{P_1}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_2}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_3}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_4}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{-P_4}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_5}{\bf{F}}\cdot{d}{\bf{r}}-\displaystyle\int{P_2}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_6}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\displaystyle\int+{P_1}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_2}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_3}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_4}{\bf{F}}\cdot{d}{\bf{r}}-\displaystyle\int{P_4}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_5}{\bf{F}}\cdot{d}{\bf{r}}-\displaystyle\int{P_2}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_6}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\displaystyle\int+{P_1}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_3}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_5}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int{P_6}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\displaystyle\int_{\partial{D}_1}{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int_{\partial{D}_2}{\bf{F}}\cdot{d}{\bf{r}} \\ &=\displaystyle\iint_{D_1}(Q_x-P_y)dA+\displaystyle\iint_{D_2}(Q_x-P_y)dA \\ &=\displaystyle\iint_{D}(Q_x-P_y)dA \end{aligned}[/latex].

Therefore, we arrive at the equation found in Green’s theorem:

[latex]\displaystyle\oint_{\partial{D}}{\bf{F}}\cdot{d}{\bf{r}}=\displaystyle\iint_D(Q_z-P_y)dA[/latex].

The same logic implies that the flux form of Green’s theorem can also be extended to a region with finitely many holes:

[latex]\displaystyle\oint_{C}{\bf{F}}\cdot{\bf{D}}ds=\displaystyle\iint_D(P_x+Q_y)dA[/latex].

Calculate integral

[latex]\large{\displaystyle\oint_{\partial{D}}\left(\sin{x}-\frac{x^3}3\right)dx+\left(\frac{y^3}3+\sin{y}\right)dy}[/latex],

where [latex]D[/latex] is the annulus given by the polar inequalities [latex]1\leq{\bf{r}}\leq2,0\leq \theta\leq2\pi[/latex].

Show Solution

Although [latex]D[/latex] is not simply connected, we can use the extended form of Green’s theorem to calculate the integral. Since the integration occurs over an annulus, we convert to polar coordinates:

[latex]\begin{aligned} \displaystyle\oint_{\partial{D}}\left(\sin{x}-\frac{x^3}3\right)dx+\left(\frac{y^3}3+\sin{y}\right)dy&=\displaystyle\iint_D(Q_x-P_y)dA \\ &=\displaystyle\iint_D(x^2+y^2)dA \\ &=\displaystyle\int_0^ {2\pi}\displaystyle\int_1^2r^3drd\theta=\displaystyle\int_0^{2\pi}\frac{15}4d\theta \\ &=\frac{15\pi}2 \end{aligned}[/latex].

Let [latex]{\bf{F}}=\langle{P},Q\rangle=\langle\frac{y}{x^2+y^2},-\frac{y}{x^2+y^2}[/latex] and let [latex]C[/latex] be any simple closed curve in a plane oriented counterclockwise. What are the possible values of [latex]\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}[/latex]?

Show Solution

We use the extended form of Green’s theorem to show that [latex]\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}[/latex] is either [latex]0[/latex] or [latex]-2\pi[/latex]–that is, no matter how crazy curve [latex]C[/latex] is, the line integral of [latex]{\bf{F}}[/latex] along [latex]C[/latex] can have only one of two possible values. We consider two cases: the case when [latex]C[/latex] encompasses the origin and the case when [latex]C[/latex] does not encompass the origin.

Case 1: [latex]C[/latex] Does Not Encompass the Origin

In this case, the region enclosed by [latex]C[/latex] is simply connected because the only hole in the domain of [latex]{\bf{F}}[/latex] is at the origin. We showed in our discussion of cross-partials that [latex]{\bf{F}}[/latex] satisfies the cross-partial condition. If we restrict the domain of [latex]{\bf{F}}[/latex] just to [latex]C[/latex] and the region it encloses, then [latex]{\bf{F}}[/latex] with this restricted domain is now defined on a simply connected domain. Since [latex]{\bf{F}}[/latex] satisfies the cross-partial property on its restricted domain, the field [latex]{\bf{F}}[/latex] is conservative on this simply connected region and hence the circulation [latex]\displaystyle\oint_C{\bf{F}}\cdot{d}{\bf{r}}[/latex] is zero.

Case 2: [latex]C[/latex] Does Encompass the Origin

In this case, the region enclosed by [latex]C[/latex] is not simply connected because this region contains a hole at the origin. Let [latex]C_1[/latex] be a circle of radius a centered at the origin so that [latex]C_1[/latex] is entirely inside the region enclosed by [latex]C[/latex] (Figure 4). Give [latex]C_1[/latex] a clockwise orientation.

Let [latex]D[/latex] be the region between [latex]C_1[/latex] and [latex]C[/latex], and [latex]C[/latex] is orientated counterclockwise. By the extended version of Green’s theorem,

[latex]\begin{aligned} \displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}+\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\iint_DQ_x-P_ydA \\ &=\displaystyle\iint_D-\frac{y^2-x^2}{(x^2+y^2)^2}+\frac{y^2-x^2}{(x^2+y^2)^2}dA \\ &=0 \end{aligned}[/latex],

and therefore

[latex]\displaystyle\int_C{\bf{F}}\cdot{d}{\bf{r}}=-\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}[/latex].

Since [latex]C_1[/latex] is a specific curve, we can evaluate [latex]\displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}[/latex]. Let

[latex]x=a\cos{t},y=a\sin{t},0\leq{t}\leq2\pi[/latex]

be a parameterization of [latex]C_1[/latex]. Then,

[latex]\begin{aligned} \displaystyle\int_{C_1}{\bf{F}}\cdot{d}{\bf{r}}&=\displaystyle\int_0^{2\pi}{\bf{F}}({\bf{r}}(t))\cdot{\bf{r}}^\prime(t)dt \\ &=\displaystyle\int_0^{2\pi}\left\langle-\frac{\sin{(t)}}a,-\frac{\cos{(t)}}1\right\rangle\cdot\langle-a\sin{(t)},-\cos{(t)}\rangle dt \\ &=\displaystyle\int_0^{2\pi}\sin^2(t)+\cos^2(t)dt=\displaystyle\int_0^{2\pi}dt=2\pi \end{aligned}[/latex].

Therefore, [latex]\displaystyle\int_C{\bf{F}}\cdot{d}s=-2\pi[/latex].